Cyclic Coordinates and Conservation of Conjugate Momenta

Before you solve a single equation of motion, read the Lagrangian the way a chess player reads a board. Every coordinate that fails to appear in $L$ — appearing only through its velocity — is a gift: its conjugate momentum $p_k = \partial L/\partial\dot q_k$ is exactly constant, forever, no integration required. Such a coordinate is called cyclic (or ignorable), and spotting one converts a second-order differential equation into a number you fix once from initial conditions. That trade — an equation for a constant — is the single most profitable move in analytical mechanics. Each conserved $p_k$ is a first integral: a relation among coordinates and velocities that holds along the whole motion. First integrals lower the order of the problem, tame it, and sometimes finish it outright; the central-force problem collapses from a plane's worth of dynamics to a one-dimensional radial equation the moment you notice the azimuth is cyclic. There is a deeper current underneath. A coordinate is absent from $L$ precisely when shifting it changes nothing — when the system has a continuous symmetry along that direction. Cyclic coordinates are symmetry made visible in your notation: translation invariance shows up as a missing $x$ and conserves linear momentum; rotational invariance as a missing $\phi$ and conserves angular momentum. But visibility depends on the coordinates you chose. The same physics hides its conservation law in Cartesian variables and displays it in polar ones. Part of the craft this article teaches is choosing coordinates so that the symmetries — and their free constants — stand in plain sight, the local preview of what Noether's theorem will make universal.

Key takeaways

A coordinate $q_k$ is cyclic (ignorable) if it does not appear explicitly in the Lagrangian — only its velocity $\dot q_k$ may. Formally: $\partial L/\partial q_k = 0$ .
The Euler–Lagrange equation then collapses to $\dot p_k = 0$ : the conjugate momentum $p_k = \partial L/\partial\dot q_k$ is a conserved first integral, fixed once by initial conditions.
What is conserved is $p_k$ , not $\dot q_k$ — in a central orbit $p_\phi = mr^2\dot\phi$ stays constant while $\dot\phi$ speeds up and slows down as $r$ changes.
Cyclicity is symmetry made visible: absence of $q_k$ means $L$ is invariant under $q_k \to q_k + \varepsilon$ , and $p_k$ is the conserved charge of that invariance — Noether's theorem in its simplest costume.
Whether a coordinate is cyclic depends on the coordinates you chose: central-force angular momentum hides in Cartesian coordinates and stands exposed in polar ones. Choosing symmetry-adapted coordinates is the art.
Each cyclic coordinate lets you eliminate one degree of freedom (Routh's procedure), leaving reduced dynamics with the conserved momentum as a fixed parameter.
Key idea: if $\partial L/\partial q_k=0$ , then $p_k=\partial L/\partial\dot q_k$ is conserved. A coordinate's absence from the Lagrangian is the visible signature of a symmetry, and the matching conjugate momentum is its conserved Noether charge.

Where this lives today

Fusion plasma physics runs on this article's theorem at industrial scale. A tokamak is engineered to be axisymmetric — its Lagrangian has a cyclic toroidal angle — so every charged particle conserves a canonical toroidal momentum $p_\varphi = mRv_\varphi + q\psi(R,Z)$ , the tokamak's exact analogue of Derivation C's $m\dot y + qBx$ (with the flux function $\psi$ playing the role of $qBx$ ). That single conserved quantity is what confines orbits to nested surfaces, and modern gyrokinetic codes preparing ITER-era experiments monitor it explicitly: numerical schemes that respect the cyclic symmetry conserve $p_\varphi$ to machine precision, while schemes that break axisymmetry leak particles for reasons this page makes obvious. The same logic now runs backwards inside machine learning. Lagrangian and Hamiltonian neural networks — models that learn $L$ or $H$ from trajectory data rather than fitting forces — inherit conservation laws structurally: if the learned Lagrangian comes out independent of some coordinate, the corresponding momentum is conserved by construction, and researchers actively engineer coordinate representations so that known symmetries appear as cyclic directions. Choosing coordinates to expose symmetry, the oldest craft in analytical mechanics, has become a regularization strategy for 2026-era scientific ML.

The physical picture

Picture the Lagrangian as a landscape over configuration space. The Euler–Lagrange equation in conjugate form, $\dot p_k = \partial L/\partial q_k$ , says that the slope of the landscape along each coordinate direction is what drives that coordinate's momentum. Where the landscape tilts, momentum changes. Where it is perfectly flat along some direction — where marching in $q_k$ changes nothing at all — there is no generalized force to push $p_k$ , and the momentum simply coasts, holding its initial value forever.

Figure

A flat direction in the Lagrangian means a coasting momentum

The two traditional names describe the two faces of this fact. Cyclic is historical: the missing coordinate is most often an angle, running around its circle without ever entering the physics. Ignorable is operational, and honest: once you know $p_k$ is a constant $\beta$ , you can solve $p_k = \partial L/\partial\dot q_k$ for $\dot q_k$ , substitute it away, and quite literally ignore the coordinate until the very end — its full history $q_k(t)$ is recovered by one quadrature after the interesting dynamics is done. The deepest face is the symmetry reading. If $q_k$ is absent from $L$ , then the transformation $q_k \to q_k + \varepsilon$ leaves the Lagrangian untouched for any $\varepsilon$ — the system cannot tell where along $q_k$ it is. Every such indifference costs the dynamics one conservation law. And because indifference is physical but notation is chosen, the same system can hide or reveal its conserved quantities depending on your coordinates:

Figure

The same orbit, two notations — polar coordinates expose what Cartesian coordinates hide

That is the practical craft: align a coordinate with each symmetry you can see. A system indifferent to horizontal position deserves a coordinate that marches horizontally; one indifferent to rotation about an axis deserves the angle about that axis. Do this, and conservation laws stop being things you prove and become things you read off.

The mechanism, in equations

Everything rests on the Euler–Lagrange equation, quoted here in the conjugate form derived in its own article. For each generalized coordinate,

\frac{\mathrm{d}}{\mathrm{d}t}\!\left(\frac{\partial L}{\partial \dot q_k}\right) - \frac{\partial L}{\partial q_k} = 0 \qquad\Longleftrightarrow\qquad \dot p_k = \frac{\partial L}{\partial q_k} \qquad (1)

where: $p_k = \partial L/\partial\dot q_k$ — the conjugate momentum, whose anatomy (mass matrices, field contributions, torque-like units) is the business of the previous article. Read (1) as a balance: the generalized force $\partial L/\partial q_k$ is the only thing that can change $p_k$ . Now the definition on which the whole article turns. A coordinate qualifies as cyclic through one derivative:

q_k \ \text{cyclic} \quad\Longleftrightarrow\quad \frac{\partial L}{\partial q_k} = 0 \qquad (2)

where: the partial derivative is taken holding the velocities fixed — it asks whether the symbol $q_k$ appears in $L$ , not whether the motion in $q_k$ matters. The velocity $\dot q_k$ may appear as prominently as it likes; indeed it must, or $p_k$ would vanish identically and the statement would be empty. Feed (2) into (1) and the conservation law falls out — no integration, no cleverness:

\frac{\partial L}{\partial q_k} = 0 \;\Longrightarrow\; \frac{\mathrm{d}p_k}{\mathrm{d}t} = 0 \;\Longrightarrow\; p_k(t) = p_k(0) \qquad (3)

where: $p_k(0)$ — a constant fixed by the initial conditions. This is a first integral of the motion: a once-differentiated statement, coordinates and velocities only, valid along the entire trajectory. Where the original equation of motion asked "how does $q_k$ accelerate?", the first integral answers "here is a quantity that never moves" — a demotion from second order to zeroth that you collect before doing any work. The flagship example deserves its own display. For a particle in any central potential, written in polar coordinates, $L = \tfrac12 m(\dot r^2 + r^2\dot\phi^2) - V(r)$ : the azimuth $\phi$ appears nowhere — only $\dot\phi$ does — so $\phi$ is cyclic and

p_\phi = \frac{\partial L}{\partial\dot\phi} = m r^2 \dot\phi = \ell = \text{const} \qquad (4)

where: $\ell$ — the angular momentum about the force centre. Equation (4) is Kepler's second law: the areal rate $\tfrac12 r^2\dot\phi = \ell/2m$ is constant, so the radius vector sweeps equal areas in equal times. Note carefully what is not constant: $\dot\phi = \ell/mr^2$ varies as the orbit breathes in and out — the comet whips around perihelion precisely so that $mr^2\dot\phi$ can stay fixed. Once a momentum is frozen, its coordinate can be evicted. The systematic tool is the Routhian, a partial Legendre transform performed only over the cyclic coordinates:

R(q_{\mathrm{nc}}, \dot q_{\mathrm{nc}}, \beta) \;=\; L - \sum_{k\,\text{cyclic}} \beta_k\,\dot q_k \qquad (5)

where: $\beta_k$ — the conserved values of the cyclic momenta, and $q_{\mathrm{nc}}$ — the surviving non-cyclic coordinates. After eliminating each $\dot q_k$ via $p_k = \beta_k$ , the function $R$ acts as an ordinary Lagrangian for the non-cyclic coordinates — a genuine reduction of the problem's dimension, with the conserved momenta demoted to fixed parameters (and the eliminated coordinates recovered at the end from $\dot q_k = -\partial R/\partial\beta_k$ ). We flag Routh's procedure here as the industrial-strength version of what Derivation B does by hand for one coordinate; its full development, and the "ignoration of coordinates" it enables in rigid-body and Hamiltonian contexts, belongs to the later chapters of this track.

Which statement answers which question

you want	use	notes
test whether $q_k$ is cyclic	$\partial L/\partial q_k \stackrel{?}{=} 0$ , eq. (2)	symbol absent from $L$ ; velocities held fixed
the conserved quantity	$p_k = \partial L/\partial\dot q_k$ , eq. (3)	the momentum, never the velocity
central-force orbit bookkeeping	$p_\phi = mr^2\dot\phi = \ell$ , eq. (4)	equal-areas law; $\dot\phi$ varies
eliminate a cyclic coordinate	Routhian, eq. (5)	reduced Lagrangian with $\beta$ as parameter
charge in a field, cyclic $y$	$p_y = m\dot y + qBx$ , eq. (7)	canonical momentum conserved, not $m\dot y$
symmetry not aligned with any single coordinate	Noether's theorem	the general machine; cyclicity is its simplest case

Derivation AWhy a Cyclic Coordinate Freezes Its Momentum

We are proving the conservation theorem (3) and, more importantly, unpacking what kind of statement it is: a first integral born from an invariance. The strategy is almost embarrassingly short — substitute the cyclic condition into the Euler–Lagrange equation — so the real work of this derivation is interpretive: verifying what the partial derivative holds fixed, connecting absence-from- $L$ to invariance under shifts, and making precise the sense in which the result lowers the order of the problem. The step to watch is the very first derivative: $\partial L/\partial q_k$ is taken at fixed velocities, and confusing it with a total derivative along the motion collapses the whole argument into nonsense.

Step 1. Write the Euler–Lagrange equation for the single coordinate $q_k$ in conjugate form:

\frac{\mathrm{d}p_k}{\mathrm{d}t} = \frac{\partial L}{\partial q_k}, \qquad p_k \equiv \frac{\partial L}{\partial\dot q_k}

Why this form: it isolates exactly the quantity whose fate we care about — the momentum — and exhibits the generalized force as its only driver.

Step 2. Impose cyclicity: the function $L(q,\dot q,t)$ contains no explicit $q_k$ , so $\partial L/\partial q_k = 0$ identically — at every point of configuration space, along any motion whatsoever, not merely on solutions. Why the distinction matters: an identity in the variables is stronger than an equation of motion; it guarantees the next step holds for the actual trajectory no matter what that trajectory does.

Step 3. Conclude $\dot p_k = 0$ and integrate trivially: $p_k(t) = p_k(0)$ for all $t$ . The function $p_k(q,\dot q)$ — in general a blend $\sum_j M_{kj}(q)\dot q_j$ of all the velocities — evaluates to the same number at every instant of the motion. Why this is called a first integral: the original system is second-order in each coordinate; this relation contains only first derivatives, so it has "already been integrated once," and it can replace one second-order equation with an algebraic relation.

Step 4. Read the same fact as an invariance. Since $q_k$ is absent, the shifted configuration $q_k \to q_k + \varepsilon$ (all else fixed) gives $L(\dots, q_k+\varepsilon, \dots) = L(\dots, q_k, \dots)$ for every $\varepsilon$ — the Lagrangian possesses a continuous one-parameter family of transformations that cost nothing. The conserved $p_k$ is precisely the charge that Noether's theorem assigns to this family; cyclicity is the special case in which the symmetry marches along a single coordinate axis, which is what makes it visible by inspection. Verified: matches Goldstein §2.6 and Landau–Lifshitz §7; for $L = \tfrac12 m\dot x^2$ the theorem returns constant $p_x = m\dot x$ — free-particle momentum conservation, as it must.

Go deeper · Derivation B · 2 minCashing In the Cyclic Azimuth: the Central Force Problem Drops a Dimension

Derivation BCashing In the Cyclic Azimuth: the Central Force Problem Drops a Dimension

The goal is to watch a first integral do real labor: starting from the planar central-force Lagrangian, we use the conserved $p_\phi$ to eliminate the angular velocity entirely, reducing two coupled degrees of freedom to a single radial problem governed by an effective potential. The strategy is the ignorable-coordinate program in miniature — conserve, solve algebraically, substitute — and it is the template for every Routhian reduction you will ever perform. The step to watch is the substitution target: eliminate $\dot\phi$ from the energy, where the centrifugal term lands with a plus sign; a common wrong turn is substituting into the Lagrangian itself, which flips that sign and wrecks the effective dynamics.

Step 1. Write the Lagrangian in symmetry-adapted coordinates:

L = \tfrac12 m\big(\dot r^2 + r^2\dot\phi^2\big) - V(r)

Why polar: the potential depends only on distance, so the rotation symmetry lines up with the single coordinate $\phi$ — which then appears nowhere in $L$ , only $\dot\phi$ does. In Cartesian coordinates the same symmetry exists but no coordinate is cyclic.

Step 2. Harvest the first integral: $\phi$ cyclic gives $p_\phi = mr^2\dot\phi = \ell$ , a constant fixed by initial data — equation (4), Kepler's equal-areas law.

Step 3. Solve the conservation law algebraically for the cyclic velocity: $\dot\phi = \ell/(mr^2)$ . Why this is legitimate: $p_\phi$ depends on $\dot\phi$ linearly with coefficient $mr^2 > 0$ , so the inversion is unique — the generic situation, guaranteed whenever the kinetic energy is positive-definite in the velocities.

Step 4. Because $L$ has no explicit time dependence, the energy $E = \tfrac12 m\dot r^2 + \tfrac12 mr^2\dot\phi^2 + V(r)$ is also conserved (the energy-function story of the next article in this chapter). Substitute $\dot\phi$ from Step 3 here:

E = \tfrac12 m\dot r^2 + \frac{\ell^2}{2mr^2} + V(r) \qquad (6)

where: the term $\ell^2/2mr^2$ — the centrifugal barrier — is the old angular kinetic energy wearing radial clothing. Equation (6) is a one-dimensional energy law in $r$ alone: the planar orbit problem has become a bead-on-a-line problem in the effective potential $V_{\mathrm{eff}}(r) = V(r) + \ell^2/2mr^2$ , whose full consequences (turning points, stable circular orbits, orbit classification) are developed in the central-force chapter of this track.

Step 5. Recover the ignored coordinate last: with $r(t)$ solved from (6), integrate $\dot\phi = \ell/mr(t)^2$ by quadrature to reconstruct $\phi(t)$ . Why nothing is lost: the cyclic coordinate was ignorable, not irrelevant — its entire history is bought back with one integral once the reduced problem is done. *Verified: matches Goldstein §3.2 and Landau–Lifshitz §14; for a circular orbit ( $\dot r = 0$ ) equation (6) reduces to $E = \ell^2/2mr^2 + V(r)$ , whose minimum over $r$ reproduces the standard circular-orbit condition $mr^3\dot\phi^2 = r^2 V'(r)$ .*

When the conserved momentum is not m times velocity

The theorem conserves $\partial L/\partial\dot q_k$ — whatever that derivative happens to be. For a charge in a magnetic field the difference between that object and $m\dot q_k$ is not pedantry; it is the physics of the orbit.

Go deeper · Derivation C · 1 minA Charged Particle with a Cyclic Coordinate: What Exactly Is Conserved

Derivation CA Charged Particle with a Cyclic Coordinate: What Exactly Is Conserved

We are deriving the conserved quantity for a charge $q$ moving in a uniform magnetic field $\mathbf{B} = B\hat{\mathbf{z}}$ , using the gauge $\mathbf{A} = (0, Bx, 0)$ in which the coordinate $y$ is cyclic — and then decoding the strange conserved object $p_y = m\dot y + qBx$ into a statement about the geometry of the orbit. The strategy: write the Lagrangian with its velocity-dependent coupling, spot the cyclic coordinate, and test the conservation law against the known circular motion. The step to watch is psychological: the conserved momentum contains a position, which feels wrong until you see what it measures — the centre of the circle.

Step 1. The Lagrangian for a charge in potentials $(\phi = 0, \mathbf{A})$ is $L = \tfrac12 m(\dot x^2 + \dot y^2) + q\,\dot{\mathbf{r}}\cdot\mathbf{A} = \tfrac12 m(\dot x^2 + \dot y^2) + qB\,x\dot y$ . Why this gauge: $\nabla\times\mathbf{A} = B\hat{\mathbf{z}}$ as required, and all the position dependence has been loaded onto $x$ — making $y$ cyclic by construction. A different gauge would hide this particular conservation law and expose another; the symmetry is physical, the bookkeeping is chosen.

Step 2. Harvest the theorem: $\partial L/\partial y = 0$ , so

p_y = \frac{\partial L}{\partial\dot y} = m\dot y + qBx = \text{const} \qquad (7)

where: the conserved object is the canonical momentum, kinetic piece plus field piece — and manifestly not $m\dot y$ , which the magnetic force changes continuously.

Step 3. Test against the known motion. The Lorentz force drives circular motion at the cyclotron frequency $\omega = qB/m$ : take $x = x_c + (v/\omega)\sin\omega t$ , $\ y = y_c + (v/\omega)\cos\omega t$ , so $\dot y = -v\sin\omega t$ . Why this is the right test: a conservation law that fails on the exact solution is no law; substituting the full orbit is the cheapest complete check available.

Step 4. Substitute into (7): $p_y = -mv\sin\omega t + qB\,x_c + qB(v/\omega)\sin\omega t$ . The oscillating pieces cancel exactly — because $qB(v/\omega) = mv$ — leaving

p_y = qB\,x_c

Why this is beautiful: the conserved canonical momentum is the $x$ -coordinate of the orbit's centre in disguise. The kinetic momentum $m\dot y$ spins around the circle; the field contribution $qBx$ counter-spins; their sum points steadfastly at the guiding centre. Conservation of $p_y$ = the statement that the circle does not drift. *Verified: $\mathrm{d}p_y/\mathrm{d}t = m\ddot y + qB\dot x = -qB\dot x + qB\dot x = 0$ using the equation of motion $m\ddot y = -qB\dot x$ ; consistent with Landau–Lifshitz, Classical Theory of Fields §21.*

Figure

The conserved canonical momentum is the orbit centre in disguise

Limiting cases

Case	Result	Meaning
translation-invariant $x$	$L = \tfrac12 m\dot x^2 + \dots$ , $\ \partial L/\partial x = 0$	$p_x = m\dot x$ conserved — linear momentum from translational symmetry
central force, azimuth $\phi$	$p_\phi = mr^2\dot\phi = \ell$	angular momentum; Kepler's equal-areas law, eq. (4)
no explicit time in $L$	$h = \sum_k p_k\dot q_k - L$ conserved	time as the "cyclic variable" — the Jacobi/energy first integral of the next article
charge with cyclic $y$ , field on	$p_y = m\dot y + qBx$ conserved	the canonical momentum freezes; $m\dot y$ alone circulates
all $n$ coordinates cyclic but one	$n-1$ constants $\beta_k$ ; Routhian is 1-DOF	maximal reduction: the whole problem collapses to a quadrature
no coordinate cyclic in your chart	nothing visible	conservation laws may still exist — rotate the chart or call Noether

Common trapIf a coordinate is cyclic, its velocity is constant

It is the momentum $p_k$ that is constant — the velocity $\dot q_k$ generally is not, and conflating them fails every orbit problem ever set. In a central orbit $p_\phi = mr^2\dot\phi$ is frozen while $\dot\phi = \ell/mr^2$ swings by orders of magnitude between aphelion and perihelion; the figure-skater pulling in her arms is the same equation with $r$ under her control. The velocity is constant only in the special case where $p_k$ depends on $\dot q_k$ alone with a constant coefficient (a lone Cartesian coordinate). Two further traps hide in the same sentence. First, cyclicity is about the coordinate's absence, not the velocity's: $\dot q_k$ must appear or there is no momentum to conserve. Second, with velocity-dependent couplings the frozen object is the full canonical $\partial L/\partial\dot q_k$ — for the cyclotron orbit, $m\dot y + qBx$ , in which the kinetic and field pieces trade back and forth while the sum stands still.

Worked examples

1 · Block sliding down a free wedge (confidence builder)

A block of mass $m = 1.0\ \mathrm{kg}$ slides down the frictionless $\theta = 30°$ face of a wedge of mass $M = 3.0\ \mathrm{kg}$ , which itself sits on a frictionless floor. Everything starts at rest. Using the wedge position $X$ and the block's slide distance $s$ along the face as coordinates, find the wedge's velocity when the block has descended $h = 0.5\ \mathrm{m}$ .

Step 1. Spot the cyclic coordinate before computing anything. The block's position is $(X + s\cos\theta,\ -s\sin\theta)$ plus constants, so $T = \tfrac12 M\dot X^2 + \tfrac12 m\big[(\dot X + \dot s\cos\theta)^2 + (\dot s\sin\theta)^2\big]$ and $V = -mgs\sin\theta$ . The coordinate $X$ appears nowhere in $L$ — only $\dot X$ does. $X$ is cyclic.

Step 2. Harvest the first integral: $p_X = \partial L/\partial\dot X = M\dot X + m(\dot X + \dot s\cos\theta) = \text{const}$ . Started from rest, the constant is zero — the system's total horizontal momentum. So $\dot X = -\,m\dot s\cos\theta/(M+m)$ : the wedge recoils, and $p_X = 0$ enforces the trade at every instant.

Step 3. Bring in energy (no explicit time in $L$ ): $mgh = \tfrac12 M\dot X^2 + \tfrac12 m\big[(\dot X + \dot s\cos\theta)^2 + (\dot s\sin\theta)^2\big]$ when the block has dropped $h$ , i.e. $s\sin\theta = h$ .

Step 4. Substitute $\dot X$ from Step 2 and solve. With $\cos 30° = 0.866$ , $\sin 30° = 0.5$ , the algebra gives $\dot s = 3.47\ \mathrm{m/s}$ , hence

\dot X = -\frac{1.0 \times 3.47 \times 0.866}{4.0} = \boxed{-0.752\ \mathrm{m/s}}

with the block moving at $(\dot X + \dot s\cos\theta,\ -\dot s\sin\theta) = (2.26,\ -1.74)\ \mathrm{m/s}$ , speed $2.85\ \mathrm{m/s}$ .

Step 5. Check both integrals: momentum $3.0\times(-0.752) + 1.0\times 2.26 = 0.00$ ✓; energy $\tfrac12(3.0)(0.752)^2 + \tfrac12(1.0)(2.85)^2 = 0.85 + 4.06 = 4.91\ \mathrm{J} = mgh$ ✓. Two first integrals, zero differential equations solved — the cyclic coordinate did all the work.

2 · Spherical pendulum: the skater effect in a cone of string (exam level)

A spherical pendulum has length $l = 1.0\ \mathrm{m}$ and bob mass $m = 0.5\ \mathrm{kg}$ . It is released at polar angle $\theta_0 = 60°$ from the downward vertical with a purely horizontal, azimuthal velocity $v_0 = 2.0\ \mathrm{m/s}$ . Find the conserved $p_\phi$ , determine whether the bob rises or falls, find the other turning angle, and compute $\dot\phi$ there.

Step 1. Lagrangian in spherical coordinates: $L = \tfrac12 m l^2(\dot\theta^2 + \sin^2\!\theta\,\dot\phi^2) + mgl\cos\theta$ . The azimuth $\phi$ is cyclic, so $p_\phi = ml^2\sin^2\!\theta\,\dot\phi$ is conserved. Initially $\dot\phi_0 = v_0/(l\sin\theta_0) = 2.0/0.866 = 2.31\ \mathrm{rad/s}$ , giving

p_\phi = 0.5 \times 1.0 \times \sin^2 60° \times 2.31 = \boxed{0.866\ \mathrm{kg\,m^2/s}}

Step 2. Rise or fall? A conical orbit at $\theta_0$ (constant $\theta$ ) requires $\dot\phi^2 = g/(l\cos\theta_0) = 9.81/0.5 = 19.6\ \mathrm{rad^2/s^2}$ ; the actual $\dot\phi_0^2 = 5.33$ falls far short. Gravity wins over the centrifugal tendency: the bob swings down and inward, toward smaller $\theta$ .

Step 3. The second first integral: $E = \tfrac12 mv_0^2 - mgl\cos\theta_0 = 1.00 - 2.45 = -1.45\ \mathrm{J}$ . At a turning angle $\dot\theta = 0$ , so, exactly as in Derivation B,

E = \frac{p_\phi^2}{2ml^2\sin^2\theta} - mgl\cos\theta

Step 4. Solve for the turning angles. $\theta = 60°$ satisfies it by construction; numerically the second root is $\theta_{\min} = \boxed{31.7°}$ . The bob oscillates in the band $31.7° \le \theta \le 60°$ , circulating all the while — the centrifugal barrier $p_\phi^2/2ml^2\sin^2\theta$ steepens as $\sin\theta$ shrinks and turns the bob around, exactly like the $\ell^2/2mr^2$ wall of equation (6).

Step 5. The skater effect at the inner turning point:

\dot\phi_{\min} = \frac{p_\phi}{ml^2\sin^2\theta_{\min}} = \frac{0.866}{0.5 \times 0.276} = \boxed{6.29\ \mathrm{rad/s}}

— the circulation rate has nearly tripled (2.31 → 6.29 rad/s, azimuthal speed 2.0 → 3.30 m/s) because $\sin^2\theta$ dropped from 0.750 to 0.276 while $p_\phi$ held exactly still. Cross-check: kinetic energy there is $\tfrac12(0.5)(3.30)^2 = 2.72\ \mathrm{J}$ and $E - V = -1.45 + 0.5\times9.81\times\cos 31.7° = 2.72\ \mathrm{J}$ ✓, with all of it azimuthal since $\dot\theta = 0$ .

Origins · 1 minJoseph-Louis Lagrange and how this was discovered

Joseph-Louis Lagrange

1736–1813 · Turin, Berlin, Paris

Lagrange built the machinery this article runs on, and he built it to avoid work — in the noblest sense. Self-taught in Turin, summoned to Berlin to fill Euler's chair, finishing in revolutionary Paris, he pursued a mechanics of pure analysis: the Mécanique analytique (1788) famously contains not one diagram. In it the ignorable coordinate appears as a systematic device — spot a coordinate the "force function" ignores, bank its conserved momentum, and reduce the problem before solving it, exactly the program of this page. His treatment of the rotating moon and the planetary secular equations leaned repeatedly on such integrals. The tradition he started — trade symmetry for constants, constants for dimensions — runs through Routh's ignoration of coordinates straight to Noether, who revealed the whole pattern as one theorem. Napoleon called him "the lofty pyramid of the mathematical sciences"; the pyramid's foundation stone is the observation that what the Lagrangian ignores, the dynamics conserves.

History

1687

— Newton's Principia derives the equal-areas law for any central force — the first conserved "momentum of rotation," a century before anyone could call it a cyclic coordinate.

1788

— Lagrange's Mécanique analytique turns absent coordinates into a method: conserved conjugate momenta as first integrals, banked before solving.

1834

— Hamilton makes

(q_k, p_k)

the fundamental pairing of dynamics; a cyclic

q_k

becomes a straight line in phase space.

1877

— Routh, in his Adams Prize essay on stability, perfects the "ignoration of coordinates": the Routhian reduction that eliminates cyclic variables wholesale.

1918

— Noether proves the general theorem: every continuous symmetry yields a conservation law — cyclic coordinates revealed as the case where the symmetry follows a single coordinate axis.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §2.6 (cyclic coordinates) and §8.3 (Routh's procedure) · Landau & Lifshitz, Mechanics, 3rd ed., §7 (momentum), §9 (angular momentum), §6 (energy) · Morin, Introduction to Classical Mechanics, Ch. 6 (conserved quantities from the Lagrangian) · Taylor, Classical Mechanics, §7.8 (ignorable coordinates and conservation)

Provenance. Definitions and the conservation theorem follow Goldstein §2.6 (cyclic coordinates) and §8.3 (Routh's procedure); the central-force reduction and angular-momentum example follow Goldstein §3.2 and Landau–Lifshitz §14. A verifier should recompute: (1) that $\partial L/\partial q_k=0$ reduces Euler–Lagrange to $\dot p_k=0$ ; (2) $p_\phi=mr^2\dot\phi$ for the polar Lagrangian; (3) the effective radial energy $E=\tfrac12 m\dot r^2+\ell^2/2mr^2+V(r)$ ; (4) the canonical momentum $p_x=m\dot x+qA_x$ from $L=\tfrac12 m\dot{\mathbf r}^2+q\dot{\mathbf r}\cdot\mathbf A-q\varphi$ . All equations are standard textbook results; no numerical constants are asserted. Worked-example numbers on this page (wedge recoil $0.752\ \mathrm{m/s}$ and block speed $2.85\ \mathrm{m/s}$ ; pendulum $p_\phi = 0.866\ \mathrm{kg\,m^2/s}$ , turning band $31.7°$ – $60°$ , $\dot\phi$ reaching $6.29\ \mathrm{rad/s}$ ) are illustrative arithmetic recomputed from the two first integrals.

Builds on: The Lagrangian Function · The Euler-Lagrange Equations · Generalized Forces and Generalized Momenta · Generalized Coordinates and Configuration Space

Leads to: Symmetries and Continuous Transformations · The Energy Function and Jacobi Integral · Degrees of Freedom

If φ is cyclic, why isn't φ̇ constant? The system "doesn't care" about φ, so why should its rate change?

The system's indifference to where it is in $\phi$ says nothing about the cost of moving in $\phi$ — that cost, the coefficient $mr^2$ in $p_\phi = mr^2\dot\phi$ , depends on the other, non-cyclic coordinates. As $r$ changes, the "exchange rate" between $\dot\phi$ and the frozen $p_\phi$ changes with it, so $\dot\phi = \ell/mr^2$ must swing to compensate. The skater is the canonical demonstration: pulling her arms in changes no symmetry and no $p_\phi$ , but it slashes the moment of inertia, and the spin rate leaps. Constant velocity is the degenerate special case where the momentum's coefficient is itself constant — a lone Cartesian coordinate with fixed mass. In the spherical-pendulum example this is worked to numbers: $p_\phi$ pinned at 0.866 kg·m²/s while $\dot\phi$ nearly triples.

How can a conservation law depend on my choice of coordinates? Physics shouldn't care about notation.

It doesn't — what depends on coordinates is visibility, not existence. The central-force problem conserves angular momentum whether you write it in Cartesian or polar variables; but in Cartesian form the invariance is a joint rotation of $x$ and $y$ , aligned with no single coordinate, so no coordinate is cyclic and the constancy of $x\dot y - y\dot x$ must be discovered by other means. Switch to polar coordinates and the symmetry marches along the $\phi$ -axis: $\phi$ vanishes from $L$ and the law reads itself off. The general tool that finds conservation laws regardless of alignment is Noether's theorem; cyclicity is its zero-effort special case, and choosing coordinates that make symmetries cyclic is how you buy theorems for free.

Is a cyclic coordinate the same thing as Noether's theorem then?

It is the simplest instance, and it is worth being precise about the gap. Cyclicity handles symmetries that translate one coordinate: $q_k \to q_k + \varepsilon$ with everything else fixed. Noether's theorem handles any continuous family of transformations that leaves the action invariant — rotations mixing several coordinates, Galilean boosts that shift velocities (where the Lagrangian changes by a total time derivative and the theorem still delivers a charge), and, in field theory, internal symmetries with no geometric picture at all. Every cyclic coordinate is a Noether symmetry whose charge is the conjugate momentum; most Noether symmetries are not cyclic in your current chart, though locally one can often choose coordinates along the symmetry flow to make them so. The two articles that follow build exactly this generalization.

Why "cyclic"? Nothing in the definition mentions circles.

The name is a fossil from the nineteenth-century problems where the phenomenon was first systematized: the absent coordinates were almost always angles — the azimuth of an orbit, the rotation angle of a top, the phase of a rotating machine — variables that run around a cycle without ever appearing in the dynamics. "Ignorable" (Routh's usage, via the phrase "ignoration of coordinates") is the operationally honest name: the coordinate can be ignored during the solution and reconstructed afterwards by a single integration, exactly as the wedge position and the pendulum azimuth are in this article's examples. Modern texts use the two words interchangeably; the definition is only ever $\partial L/\partial q_k = 0$ , circles or not.

Time never appears in many Lagrangians. Is time a cyclic coordinate, and is energy its momentum?

Morally yes, technically with a wrinkle worth respecting. Time is not a generalized coordinate — it is the parameter of the motion — so the theorem of this article does not literally apply. But the analogous statement holds: when $\partial L/\partial t = 0$ , the energy function $h = \sum_k p_k\dot q_k - L$ is conserved, and its derivation (differentiate $L$ along the motion, use the Euler–Lagrange equations) parallels the cyclic argument step for step. The object $h$ , when and why it equals the mechanical energy $T + V$ , and the spectacular cases where it is conserved while $T + V$ is not, are the entire subject of the next article. In the Hamiltonian picture the analogy becomes exact: time and (minus) energy form a conjugate pair on the same footing as $q$ and $p$ .

For the charged particle, the conserved p_y contains the position x. How can a "momentum" depend on where the particle is?

Because the theorem conserves $\partial L/\partial\dot y$ , and with the velocity-dependent coupling $qB\,x\dot y$ in the Lagrangian that derivative is $m\dot y + qBx$ — the canonical momentum, which inherits a position term from the field. The physical decoding is the satisfying part: along the exact cyclotron orbit, the kinetic piece $m\dot y$ oscillates and the field piece $qBx$ oscillates in perfect anti-phase, and their sum equals $qB\,x_c$ — a constant proportional to the orbit centre's coordinate. Conservation of $p_y$ is the statement that the guiding centre does not drift. Plasma physicists build entire confinement devices on the generalization of this fact: a tokamak's conserved canonical toroidal momentum is what ties particle orbits to magnetic surfaces.

Can I always find coordinates in which a given conserved quantity becomes cyclic?

Locally, yes — if you already know the symmetry. Any one-parameter continuous symmetry defines a flow in configuration space, and coordinates can be chosen so that one of them marches along that flow; in those coordinates the Lagrangian loses the marching coordinate and the conserved quantity becomes its conjugate momentum. (For central forces this is precisely the passage to polar coordinates.) The catches are real, though: finding the symmetry in the first place is the hard problem, several symmetries at once may not fit a single coordinate grid (the three components of angular momentum cannot all be cyclic simultaneously — their symmetries do not commute), and chaotic systems typically have no conserved quantities beyond energy, hence no such coordinates at all. Where the program does succeed completely, it terminates in action-angle variables — the Hamiltonian chapters of this track.

Does each cyclic coordinate really remove a full degree of freedom, or does something remain?

Each cyclic coordinate removes one degree of freedom from the active problem, with two honest footnotes. First, the eliminated coordinate leaves a residue: its conserved momentum enters the reduced dynamics as a parameter, typically as a barrier term like $\ell^2/2mr^2$ — the reduced problem is genuinely one dimension smaller but "remembers" the circulation through its effective potential. Second, the coordinate itself is postponed, not destroyed: after solving the reduced system you reconstruct $q_k(t)$ by one quadrature, as in Step 5 of Derivation B. With several cyclic coordinates the reductions stack — the heavy symmetric top has two (precession and spin angles), collapsing three rotational degrees of freedom to a single equation in the tilt angle, which is why that classic problem is solvable at all.

Cyclic Coordinates and Conservation of Conjugate Momenta

Where this lives today

The physical picture

The mechanism, in equations

Which statement answers which question

When the conserved momentum is not m times velocity

Limiting cases

Worked examples

1 · Block sliding down a free wedge (confidence builder)

2 · Spherical pendulum: the skater effect in a cone of string (exam level)

Going further

Before this

Where this leads