Cyclic Coordinates and Conserved Momenta

Write the central-force problem in Cartesian coordinates and the Hamiltonian is an honest mess: $x$ and $y$ both appear, tangled inside $V(\sqrt{x^2+y^2})$ , and nothing is obviously conserved. Write the same problem in polar coordinates and the angle $\theta$ simply vanishes from $H$ . Hamilton's equation $\dot p_\theta=-\partial H/\partial\theta$ then reads $\dot p_\theta=0$ before you have solved anything: angular momentum is conserved, and a two-dimensional problem collapses to one radial equation plus a final integral. A coordinate that is absent from the Hamiltonian in this way is called cyclic (or ignorable), and it is the single cheapest conservation law in physics — you get it by inspection. This page is the Hamiltonian half of the story, and the half with the sharper payoff. You have already seen (in Cyclic Coordinates and Conservation of Conjugate Momenta) that a coordinate missing from the Lagrangian conserves its conjugate momentum; here the same fact becomes a genuine reduction: the conserved momentum turns into a mere number in the reduced Hamiltonian, the cyclic pair drops out of the active dynamics entirely, and Routh's hybrid construction packages the whole procedure into an effective one-dimensional problem. But the deepest lesson is a provocation. Whether a coordinate is cyclic is a property of your coordinate choice, not of the system — the same orbit hides its conservation law in Cartesian variables and displays it in polar ones. So why stop at one good coordinate? The dream of making every coordinate cyclic is precisely what launches the theory of canonical transformations, the chapter this article opens the door to.

Key takeaways

A coordinate $q_k$ is cyclic (ignorable) if it does not appear explicitly in the Hamiltonian — even though its conjugate momentum $p_k$ usually does.
Hamilton's equation $\dot p_k=-\partial H/\partial q_k$ then gives $\dot p_k=0$ immediately: $p_k$ is a constant of the motion, fixed by the initial conditions.
The Hamiltonian reduction is complete: the conserved $p_k=\alpha$ enters the reduced Hamiltonian as a parameter, the active phase space shrinks from $2n$ to $2(n-1)$ dimensions, and $q_k(t)$ is recovered at the end by one quadrature.
On the Lagrangian side the reduction is not automatic — the cyclic velocity $\dot q_k$ still appears in $L$ ; the Routhian $R=\sum_{\text{cyc}}p_j\dot q_j-L$ fixes this by Legendre-transforming only the cyclic variables.
Cyclicity depends on the choice of coordinates: central forces have no cyclic Cartesian coordinate but a cyclic polar angle. Choosing — or transforming to — coordinates adapted to the symmetry is the whole game.
Key idea: If $\partial H/\partial q_k = 0$ then $p_k$ is conserved; the cyclic coordinate $q_k$ drops out of the active dynamics and is recovered at the end by one integration.

Where this lives today

The purest industrial descendant of the cyclic coordinate is guiding-center and gyrokinetic theory in magnetized plasmas. A charged particle in a tokamak gyrates around a field line millions of times per transit; the gyrophase is (to excellent approximation) an ignorable angle, and its conjugate momentum — the magnetic moment $\mu$ — is the frozen quantity that makes the simulation tractable. Gyrokinetic codes used to design fusion experiments like ITER are, at bottom, Routh's maneuver executed asymptotically: eliminate the fast cyclic angle, keep its conserved momentum as a parameter, and evolve only the slow remaining dynamics — a million-fold saving in time resolution. The same reduction, in its modern geometric dress (symplectic reduction), runs through spacecraft attitude control, the falling-cat problem and robot locomotion — where the "cyclic" directions are overall rotations whose conserved angular momentum constrains what shape-changes can achieve — and through molecular-dynamics codes, which exploit conserved total momentum to factor out the center-of-mass drift of every simulation box. Whenever a simulation "works in the co-moving/co-rotating frame with fixed angular momentum," a cyclic coordinate has been ignored, in Routh's precise sense of the word.

The physical picture

Think of a cyclic coordinate as a dial you can turn without changing the machine's internal state. Nothing in the energy bookkeeping notices the dial's setting, so whatever momentum is associated with turning it has nothing to push against — it stays frozen at its initial value. A frictionless puck on an infinite featureless table has two such dials ( $x$ and $y$ : linear momentum conserved); a spinning axisymmetric top has its spin angle ( $p_\psi$ conserved); a planet orbiting the Sun has the polar angle ( $p_\theta=\ell$ conserved). The crucial subtlety is that the same physical system shows or hides its dials depending on the coordinates you draw. That is the hero picture:

Figure

One orbit, two coordinate charts — Cartesian coordinates hide the conservation law that polar coordinates hand you by inspection

Geometrically, each conserved $p_k$ traps the trajectory on a level sheet $p_k=\alpha$ inside the $2n$ -dimensional phase space. The motion never leaves that sheet, so the active dynamics — the part you must actually integrate — lives in two fewer dimensions. Stack up one cyclic coordinate after another and the problem cascades downward:

Figure

The reduction cascade — every cyclic pair freezes onto a sheet of phase space, leaving fewer active equations and one quadrature IOU

The bookkeeping is hierarchical, and worth internalizing as a recipe: solve the reduced problem for the non-cyclic variables first, treating each conserved momentum as a number; only then integrate $\dot q_k=\partial H/\partial p_k$ — now an explicit function of time — to recover where the ignorable coordinate actually went. The ignorable coordinate is not unphysical; it is simply deferrable.

Cyclic coordinates in the Hamiltonian language

Everything starts from a definition you can check by inspection. Given a Hamiltonian $H(q,p,t)$ — built, as always, by the Legendre transformation from $L$ — scan its argument list:

\frac{\partial H}{\partial q_k}=0 \;\Longleftrightarrow\; q_k\ \text{is cyclic} \qquad (1)

where: $H$ — the Hamiltonian $H(q,p,t)$ ; $q_k$ — the coordinate under test; $p_k$ — its conjugate momentum, which may (and usually does) still appear in $H$ . Read it physically: the energy landscape is flat along the $q_k$ direction. Note what is not required — $p_k$ absent, $\dot q_k$ absent, or $H$ time-independent. Only the coordinate itself must be missing. The conservation law is then a one-line consequence of Hamilton's canonical equations (see Hamilton's Canonical Equations — we use them here, not re-derive them). The momentum equation $\dot p_k=-\partial H/\partial q_k$ collapses:

\dot p_k=-\frac{\partial H}{\partial q_k}=0 \;\;\Longrightarrow\;\; p_k=\alpha=\text{const} \qquad (2)

where: $\alpha$ — the frozen value of $p_k$ , set once by the initial conditions. This is the entire proof — no clever integrating factor, no Noether machinery. The Hamiltonian formalism was built so that conservation laws would read off like this: one flat direction in $H$ , one frozen momentum. Whether that frozen number is a linear momentum, an angular momentum, or something less familiar (the canonical $p_\phi$ of a charged particle includes a field term) depends on the coordinate, not on the theorem.

Derivation ACyclic implies conserved, and the problem genuinely drops a dimension

We are proving the two claims that make cyclic coordinates the workhorse of Hamiltonian problem-solving: that $p_k$ is conserved the moment $q_k$ is absent from $H$ , and — the part the Lagrangian formulation cannot match — that the cyclic pair $(q_k,p_k)$ then leaves the active dynamics entirely, reducing an $n$ -degree-of-freedom problem to $n-1$ degrees of freedom plus one final integration. The strategy is to read Hamilton's equations with the flat direction imposed, check that cyclicity is a property of the physics and not of the formalism by carrying it through the Legendre transform, and then count what is left. Watch the contrast with the Lagrangian side: there the conserved quantity exists but the cyclic velocity $\dot q_k$ still contaminates the remaining equations.

Step 1. Impose the flat direction on the momentum equation. For every index, Hamilton's equations read $\dot q_i=\partial H/\partial p_i$ and $\dot p_i=-\partial H/\partial q_i$ . If $\partial H/\partial q_k=0$ , then $\dot p_k=0$ , so $p_k(t)=p_k(0)\equiv\alpha$ for all time.

Step 2. Check that cyclicity transfers through the Legendre transformation. Differentiating $H=\sum_j p_j\dot q_j-L$ at fixed $(p,t)$ , with $\dot q_j=\dot q_j(q,p,t)$ implicit, the chain rule gives

\frac{\partial H}{\partial q_k}=\sum_j p_j\frac{\partial \dot q_j}{\partial q_k}-\frac{\partial L}{\partial q_k}-\sum_j\frac{\partial L}{\partial \dot q_j}\frac{\partial \dot q_j}{\partial q_k}=-\frac{\partial L}{\partial q_k}

because $p_j=\partial L/\partial\dot q_j$ makes the two sums cancel identically. Why this matters: $q_k$ is cyclic in $H$ exactly when it is cyclic in $L$ — the property belongs to the system-plus-chart, not to which formalism you opened. The conserved quantity here is the same one the Lagrangian sibling article finds via $\mathrm{d}(\partial L/\partial\dot q_k)/\mathrm{d}t=0$ .

Step 3. Substitute $p_k=\alpha$ into the Hamiltonian. Since $q_k$ never appears and $p_k$ appears only as the constant $\alpha$ , the function

H_{\text{red}}=H(q_1,\dots,q_{n-1},\,p_1,\dots,p_{n-1};\,\alpha,\,t)

generates the complete dynamics of the remaining $n-1$ pairs: their $2(n-1)$ canonical equations close among themselves, with $\alpha$ playing the role of a fixed parameter like a mass or a coupling constant. Why the Lagrangian cannot do this directly: setting $\partial L/\partial q_k=0$ conserves $p_k=\partial L/\partial\dot q_k$ , but this is a relation among the velocities — $\dot q_k$ itself still appears in $L$ and hence in every remaining Euler–Lagrange equation. To exploit the constant you must solve for $\dot q_k$ and eliminate it by hand, and doing that carelessly inside $L$ produces wrong equations of motion. The cure is Routh's construction, derived next.

Step 4. Recover the ignorable coordinate by quadrature. Along the solved reduced trajectory, $\dot q_k=\partial H/\partial p_k\big|_{p_k=\alpha}$ is a known function of time, so

q_k(t)=q_k(0)+\int_0^t \frac{\partial H}{\partial p_k}\,\mathrm{d}t'

is a single integration — no differential equation remains for $q_k$ .

Which statement when

you need	reach for	notes
spot a conserved momentum by inspection	eq. (1)	scan $H$ for missing coordinates
the conservation law itself	eq. (2)	$p_k=\alpha$ , one line from Hamilton's equations
the reduced problem	$H_{\text{red}}(q,p;\alpha,t)$	cyclic pair gone; $\alpha$ a parameter
the ignorable coordinate's history	quadrature, Step 4 above	do it last
reduction on the Lagrangian side	Routhian, eq. (3)	half Legendre transform
effective 1D dynamics for central forces	Derivation C	$V_{\text{eff}}=\ell^2/2mr^2+V(r)$

The Routhian: reduce like a Hamiltonian, solve like a Lagrangian

Sometimes you want the cyclic coordinates eliminated and the surviving equations in familiar Euler–Lagrange form — for instance to reuse small-oscillation machinery about a steady motion. Routh's construction does exactly this: perform the Legendre transformation only over the cyclic variables, leaving the honest coordinates in Lagrangian dress. Suppose $q_{s+1},\dots,q_n$ are cyclic. Define

R(q_1,\dots,q_s,\,\dot q_1,\dots,\dot q_s,\,p_{s+1},\dots,p_n,\,t) \;=\; \sum_{j=s+1}^{n} p_j\,\dot q_j \;-\; L \qquad (3)

where: $R$ — the Routhian, a hybrid of $H$ and $-L$ ; the sum runs over the cyclic coordinates only, whose velocities are eliminated in favor of the conserved momenta $p_j=\beta_j$ ; the non-cyclic $\dot q_i$ survive untouched. Think of $R$ as a partial Legendre transform (the full-transform machinery is in Legendre Transformation): Hamiltonian where that pays (the frozen directions), Lagrangian where that is comfortable (the live ones). The dynamics generated by $R$ splits cleanly along the same line:

\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial R}{\partial \dot q_i}-\frac{\partial R}{\partial q_i}=0 \;\;(i\le s), \qquad \dot q_j=\frac{\partial R}{\partial p_j},\;\; \dot p_j=-\frac{\partial R}{\partial q_j}=0 \;\;(j>s) \qquad (4)

where: the first set are Euler–Lagrange equations with $R$ standing in for the Lagrangian of the reduced system; the second set are Hamilton's equations for the cyclic pairs, the momentum ones confirming $p_j=\beta_j$ . In practice you use only the first set — with every $\beta_j$ inserted as a number — and pay the quadrature IOU for the $q_j$ at the end.

Go deeper · Derivation B · 1 minRouth's hybrid function and its split equations of motion

Derivation BRouth's hybrid function and its split equations of motion

The goal is to justify equation (4): that after the partial Legendre transform (3), the non-cyclic coordinates obey Euler–Lagrange equations generated by $R$ , while the cyclic pairs obey Hamilton's equations — so the conserved momenta can be legally substituted as constants before deriving the reduced equations of motion. The strategy is the standard differential bookkeeping of any Legendre transform: expand $\mathrm{d}R$ from the definition, cancel the conjugate-pair terms, and read off partial derivatives. The point to watch is which variables $R$ actually depends on — that is the entire content of the construction.

Step 1. Take the total differential of definition (3), using $\mathrm{d}L=\sum_i\big(\tfrac{\partial L}{\partial q_i}\mathrm{d}q_i+\tfrac{\partial L}{\partial\dot q_i}\mathrm{d}\dot q_i\big)+\tfrac{\partial L}{\partial t}\mathrm{d}t$ over all $n$ coordinates:

\mathrm{d}R=\sum_{j>s}\big(\dot q_j\,\mathrm{d}p_j+p_j\,\mathrm{d}\dot q_j\big)-\sum_{i}\frac{\partial L}{\partial q_i}\,\mathrm{d}q_i-\sum_{i}\frac{\partial L}{\partial \dot q_i}\,\mathrm{d}\dot q_i-\frac{\partial L}{\partial t}\,\mathrm{d}t

Step 2. Cancel the cyclic velocity differentials. For $j>s$ , the definition of momentum gives $\partial L/\partial\dot q_j=p_j$ , so the $+p_j\,\mathrm{d}\dot q_j$ and $-p_j\,\mathrm{d}\dot q_j$ terms annihilate. Moreover $\partial L/\partial q_j=0$ because those coordinates are cyclic. What survives is

\mathrm{d}R=\sum_{j>s}\dot q_j\,\mathrm{d}p_j-\sum_{i\le s}\Big(\frac{\partial L}{\partial q_i}\,\mathrm{d}q_i+\frac{\partial L}{\partial \dot q_i}\,\mathrm{d}\dot q_i\Big)-\frac{\partial L}{\partial t}\,\mathrm{d}t

Why the cancellation is the construction: no $\mathrm{d}\dot q_j$ survives for the cyclic directions, so $R$ genuinely depends on $(q_i,\dot q_i)$ for $i\le s$ and on $p_j$ for $j>s$ — precisely the hybrid argument list claimed in (3).

Step 3. Read off coefficients: $\partial R/\partial p_j=\dot q_j$ and $\partial R/\partial q_j=0=-\dot p_j$ for the cyclic pairs (Hamilton's equations, conservation included), and $\partial R/\partial q_i=-\partial L/\partial q_i$ , $\partial R/\partial\dot q_i=-\partial L/\partial\dot q_i$ for the live coordinates. Substituting the last two into the Euler–Lagrange equation for $q_i$ flips both signs at once, leaving its form intact with $R$ in place of $L$ — which is (4). Why $-R$ is the honest effective Lagrangian: since both partials of $R$ carry a minus sign relative to $L$ , the function $L_{\text{eff}}=-R$ has the standard signs. For the central force this is where the centrifugal term famously flips: the kinetic piece $\tfrac{\ell^2}{2mr^2}$ of $L$ reappears inside $-R$ as a repulsive potential $+\tfrac{\ell^2}{2mr^2}$ in $V_{\text{eff}}$ — forget the flip and steady radii come out wrong.

Reduction in practice: the central force, start to finish

The planar central-force problem is the canonical demonstration — two degrees of freedom in, one out, and every step of the cascade visible. It is also the computation behind Kepler's second law and behind Worked Example 1 below.

Go deeper · Derivation C · 1 minCentral force reduced to one radial degree of freedom by the cyclic angle

Derivation CCentral force reduced to one radial degree of freedom by the cyclic angle

We are executing the full reduction pipeline on a particle of mass $m$ in a central potential $V(r)$ : identify the cyclic coordinate, freeze its momentum, build the reduced one-dimensional Hamiltonian with an effective potential, and end with the quadrature that recovers the angle. The strategy is simply Derivation A run on a concrete system, so each abstract step acquires a face. Watch how the conserved $\ell$ migrates from "momentum" to "parameter": the term $p_\theta^2/2mr^2$ , kinetic in origin, becomes a repulsive $\ell^2/2mr^2$ barrier in the radial problem.

Step 1. Lagrangian and momenta in plane polar coordinates. From $L=\tfrac12 m(\dot r^2+r^2\dot\theta^2)-V(r)$ , the conjugate momenta are $p_r=m\dot r$ and $p_\theta=mr^2\dot\theta$ — the angular momentum about the force center.

Step 2. Build the Hamiltonian by the (full) Legendre transform and scan it:

H=\frac{p_r^2}{2m}+\frac{p_\theta^2}{2mr^2}+V(r)

The angle $\theta$ is nowhere in sight: it is cyclic, by inspection, exactly as promised by the hero figure.

Step 3. Freeze the conserved momentum: $\dot p_\theta=-\partial H/\partial\theta=0$ , so $p_\theta=\ell$ , a number fixed by the launch conditions. Why this is rotational symmetry speaking: $\theta$ absent from $H$ is the statement that the physics is invariant under rotations about the center — the symmetry-side of this coin is told properly in the Lagrangian sibling article and in Noether's theorem, and we do not retell it here.

Step 4. Read off the reduced one-dimensional problem. With $\ell$ inserted, the radial pair obeys the Hamiltonian

H_{\text{red}}=\frac{p_r^2}{2m}+\underbrace{\frac{\ell^2}{2mr^2}+V(r)}_{V_{\text{eff}}(r)}

a bona fide 1D system: $\dot r=p_r/m$ , $\dot p_r=-V_{\text{eff}}'(r)$ . Turning points, stability of circular orbits, oscillation frequencies — all now read off a one-dimensional potential. Why the barrier term is physical: $\ell^2/2mr^2$ is the kinetic energy the particle is obliged to keep because angular momentum cannot leak away; shrinking $r$ at fixed $\ell$ forces faster rotation. It is not a new force — it is the old conservation law wearing a potential's clothes.

Step 5. Pay the quadrature IOU. Once $r(t)$ is solved, the angle follows from one integration:

\theta(t)=\theta_0+\int_0^t \frac{\ell}{m\,r(t')^2}\,\mathrm{d}t'

and the areal-velocity statement $\tfrac12 r^2\dot\theta=\ell/2m=\text{const}$ is Kepler's second law, free of charge.

The canonical-transformation dream

Here is the thought that turns a bookkeeping trick into a research program. Cyclicity is chart-dependent: the central force has zero cyclic Cartesian coordinates but one cyclic polar coordinate. Choosing better coordinates created a conserved momentum out of thin air. Push the greed to its logical end: could there be coordinates in which every coordinate is cyclic? Then $H=H(P_1,\dots,P_n)$ alone, every $P_i$ is constant, every $Q_i$ evolves linearly in time — the system is solved by pure algebra:

H=H(P) \;\;\Longrightarrow\;\; \dot P_i=0,\qquad \dot Q_i=\frac{\partial H}{\partial P_i}=\text{const} \qquad (5)

where: $(Q,P)$ — a hypothetical set of phase-space coordinates adapted so perfectly to the dynamics that nothing accelerates. Reaching them requires transformations that mix coordinates and momenta wholesale — far beyond the point transformations $Q=Q(q,t)$ of Lagrangian mechanics — while preserving the canonical form of Hamilton's equations. Which changes of variables are allowed, and how to build them, is precisely the subject of the next article, Definition of Canonical Transformations; that the goal is achievable for a large class of systems is the triumph of Hamilton–Jacobi theory and action-angle variables.

Figure

From zero cyclic coordinates to all of them — the road that motivates the canonical-transformations chapter

Limiting cases

Case	Result	Meaning
free particle, Cartesian chart	$H=(p_x^2+p_y^2+p_z^2)/2m$	all three coordinates cyclic; full momentum vector frozen — maximal reduction, motion solved by inspection
central force, polar chart	$H=p_r^2/2m+p_\theta^2/2mr^2+V(r)$	one cyclic angle; $p_\theta=\ell$ frozen, radial 1D problem with $V_{\text{eff}}$ remains
central force, Cartesian chart	no coordinate cyclic	same conserved $\ell=xp_y-yp_x$ exists but is invisible to inspection — cyclicity is chart-dependent
heavy symmetric top	$\phi,\psi$ cyclic in $H(\theta,p_\theta,p_\phi,p_\psi)$	two frozen momenta; only nutation $\theta$ is dynamical — a 3-DOF problem collapses to 1
$H$ explicitly time-dependent, $q_k$ absent	$p_k$ still conserved	cyclic conservation survives when energy conservation dies
all coordinates cyclic	$H=H(P)$ : $\dot P=0$ , $\dot Q=\text{const}$	the canonical-transformation dream (5); realized by Hamilton–Jacobi / action-angle variables

Common trapCyclic means the coordinate is absent — not its momentum, and not in every chart

Two misreadings account for most lost marks. First: cyclicity requires $q_k$ absent from $H$ ; the conjugate $p_k$ almost always still appears (in the central force, $p_\theta^2/2mr^2$ sits in plain sight while $\theta$ is cyclic). Do not disqualify a coordinate because its momentum shows up. Second: cyclicity is a property of the chart, not the system. The same central-force problem has a cyclic coordinate in polar variables and none in Cartesian variables, yet angular momentum is conserved either way — in the bad chart the constant is the invisible combination $xp_y-yp_x$ . Corollary for problem-solving: if you expect a symmetry, pick (or transform to) coordinates along it before building $H$ ; and never "simplify" by substituting a conserved momentum back into the Lagrangian — that mangles the remaining equations. Eliminate cyclic velocities only through the Routhian (3) or work in the Hamiltonian where substitution of $p_k=\alpha$ is legal.

Worked examples

1 · Satellite in Earth's gravity: freeze the angle, read the orbit (confidence builder)

A satellite of mass $m=800\ \mathrm{kg}$ is at $r_0=7000\ \mathrm{km}$ from Earth's center, moving purely tangentially at $v_0=7.50\ \mathrm{km/s}$ . Take $GM_\oplus=3.986\times10^{14}\ \mathrm{m^3/s^2}$ . Find the conserved $p_\theta$ , the effective potential picture at $r_0$ , and the other apsis of the orbit.

Step 1. Identify the cyclic coordinate and freeze its momentum. In the plane of motion, $H=p_r^2/2m+p_\theta^2/2mr^2-GMm/r$ has $\theta$ cyclic, so $p_\theta=\ell=mr_0v_0=800\times(7.0\times10^{6})\times7500=\mathbf{4.20\times10^{13}\ kg\,m^2/s}$ — locked for the whole flight.

Step 2. Evaluate the reduced 1D energy. The launch is tangential, so $p_r=0$ and $r_0$ is an apsis. Centrifugal term: $\ell^2/2mr_0^2=\tfrac12 mv_0^2=\tfrac12\times800\times(7500)^2=2.25\times10^{10}\ \mathrm{J}$ . Gravity: $-GMm/r_0=-3.1888\times10^{17}/7.0\times10^{6}=-4.555\times10^{10}\ \mathrm{J}$ . Total: $E=V_{\text{eff}}(r_0)=-2.305\times10^{10}\ \mathrm{J}<0$ — bound orbit.

Step 3. Which way does $r$ move first? $V_{\text{eff}}'(r_0)=-\ell^2/mr_0^3+GMm/r_0^2=-6429+6508=+79\ \mathrm{N}$ , so the radial force $-V_{\text{eff}}'=-79\ \mathrm{N}$ points inward: the satellite falls toward Earth initially. (Cross-check: circular speed at $r_0$ is $\sqrt{GM/r_0}=7.546\ \mathrm{km/s}$ ; we launched slower, so $r_0$ must be the apogee.)

Step 4. Find the perigee from the frozen quantities alone. Semi-major axis from the vis-viva energy: $a=-GMm/2E=3.1888\times10^{17}/(4.611\times10^{10})=6.916\times10^{6}\ \mathrm{m}$ , so $r_p=2a-r_0=6.832\times10^{6}\ \mathrm{m}$ (perigee altitude $\approx 454\ \mathrm{km}$ ). Perigee speed from the conserved $p_\theta$ : $v_p=\ell/mr_p=4.20\times10^{13}/(800\times6.832\times10^{6})=\mathbf{7.685\ km/s}$ . Energy check there: $\tfrac12mv_p^2-GMm/r_p=2.362\times10^{10}-4.667\times10^{10}=-2.305\times10^{10}\ \mathrm{J}$ ✓ — same $E$ , as conservation demands.

2 · Spherical pendulum: one cyclic angle, one live angle (exam level)

A pendulum bob of mass $m=0.20\ \mathrm{kg}$ swings on a rigid massless rod of length $l=1.0\ \mathrm{m}$ . At $t=0$ it is at polar angle $\theta_0=60^\circ$ from the downward vertical with $\dot\theta_0=0$ and azimuthal rate $\dot\phi_0=3.0\ \mathrm{rad/s}$ . Take $g=9.8\ \mathrm{m/s^2}$ . Find the conserved momentum, decide whether the bob initially rises or falls, and compute the initial $\ddot\theta$ .

Step 1. Scan for the cyclic coordinate. With $L=\tfrac12ml^2(\dot\theta^2+\sin^2\!\theta\,\dot\phi^2)+mgl\cos\theta$ , the azimuth $\phi$ is absent: cyclic. Its frozen momentum is $p_\phi=ml^2\sin^2\!\theta\,\dot\phi=0.20\times1.0\times(0.8660)^2\times3.0=0.20\times0.75\times3.0=\mathbf{0.450\ kg\,m^2/s}$ .

Step 2. Build the reduced problem for the live angle. Eliminating $\dot\phi$ via the Routhian ( $L_{\text{eff}}=-R$ ) gives a 1D system in $\theta$ with $U_{\text{eff}}(\theta)=p_\phi^2/(2ml^2\sin^2\!\theta)-mgl\cos\theta$ — note the sign flip that turns the azimuthal kinetic term into a repulsive barrier around the vertical.

Step 3. Compare against steady (conical) motion. A horizontal circle at $\theta_0=60^\circ$ would need $\dot\phi_c=\sqrt{g/(l\cos\theta_0)}=\sqrt{9.8/0.50}=4.43\ \mathrm{rad/s}$ . Our bob spins at only $3.0\ \mathrm{rad/s}$ — too slow for the centrifugal barrier to hold it out at $60^\circ$ , so gravity wins and the bob initially falls toward the vertical ( $\theta$ decreases... careful: $\theta$ is measured from the downward vertical, so falling toward the vertical means $\theta$ decreases).

Step 4. Compute the initial angular acceleration from $ml^2\ddot\theta=-U_{\text{eff}}'(\theta)$ . The barrier term: $p_\phi^2\cos\theta_0/(ml^2\sin^3\!\theta_0)=(0.450)^2\times0.50/(0.20\times0.6495)=0.1013/0.1299=0.779\ \mathrm{N\,m}$ . The gravity term: $mgl\sin\theta_0=0.20\times9.8\times0.8660=1.697\ \mathrm{N\,m}$ . So $ml^2\ddot\theta=0.779-1.697=-0.918\ \mathrm{N\,m}$ and $\ddot\theta_0=-0.918/0.20=\mathbf{-4.59\ rad/s^2}$ — consistent with Step 3's verdict. The motion oscillates in $\theta$ between two turning circles while $\phi$ advances by the quadrature $\dot\phi=p_\phi/ml^2\sin^2\!\theta$ , fastest when the bob dips closest to the vertical — the same $1/\!\sin^2\theta$ speedup a Kepler orbit shows at perihelion via $1/r^2$ .

Origins · 1 minEdward Routh and how this was discovered

Edward Routh

1831–1907 · Cambridge

Edward Routh was Victorian Cambridge's supreme coach: over three decades he trained more than 600 students for the Mathematical Tripos, among them 27 Senior Wranglers — J. J. Thomson, Rayleigh, and Larmor passed through his drilling. Beating the young James Clerk Maxwell to Senior Wrangler in 1854 made his name; the mathematics that kept it was his 1877 Adams Prize essay on the stability of motion, which introduced both the Routh–Hurwitz stability criterion of modern control theory and the "ignoration of coordinates" — the Routhian — that eliminates cyclic variables before a stability analysis. That combination was no accident: to decide whether a spinning system (a governor, a top, a bicycle) is stable, one first strips away the uniform spinning via the cyclic coordinates, then examines small oscillations of what remains. Routh's procedure is still how one studies steady motions, from satellite attitude dynamics to the gaits of walking robots.

History

1788

— Lagrange's Mécanique analytique shows coordinates absent from the dynamics yield first integrals

1834

— Hamilton's canonical equations make cyclic-implies-conserved a one-line inspection

1877

— Routh's Adams Prize essay introduces the Routhian and the ignoration of coordinates for stability analysis

1904

— Whittaker's Analytical Dynamics systematizes reduction by ignorable coordinates and quadratures

1918

— Noether's theorem generalizes the coordinate-shift symmetry behind cyclicity to arbitrary continuous symmetries

1974

— Marsden and Weinstein recast the cyclic-pair dimension drop as symplectic reduction, the modern geometric form

Going further

References. Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §2.6, §8.2–8.3 · Landau & Lifshitz, Mechanics, 3rd ed., §14, §40–41 · Morin, Introduction to Classical Mechanics, Ch. 6 · Taylor, Classical Mechanics, §7.8 and §13.4 · Whittaker, Analytical Dynamics, Ch. III (ignoration of coordinates) The key identities used throughout are: the Legendre-transform relation $\partial H/\partial q_k=-\partial L/\partial q_k$ (derived by the chain-rule argument in the derivation above); the central-force momenta $p_\theta=mr^2\dot\theta$ and effective potential $\ell^2/2mr^2+V(r)$ ; and the symmetric-top Hamiltonian with its two cyclic angles. The worked-example numbers are illustrative arithmetic computed directly from the displayed formulas: satellite ( $\ell=4.20\times10^{13}\ \mathrm{kg\,m^2/s}$ , $E=-2.305\times10^{10}\ \mathrm{J}$ , $r_p=6.832\times10^{6}\ \mathrm{m}$ , $v_p=7.685\ \mathrm{km/s}$ ); spherical pendulum ( $p_\phi=0.450\ \mathrm{kg\,m^2/s}$ , $\dot\phi_c=4.43\ \mathrm{rad/s}$ , $\ddot\theta_0=-4.59\ \mathrm{rad/s^2}$ ).

Builds on: The Hamiltonian Function · Generalized Momenta · Hamilton's Canonical Equations · Legendre Transformation Leads to: Definition of Canonical Transformations · Phase Space · Hamiltonian Flow and the Phase Fluid · Equivalence of Lagrangian and Hamiltonian Formulations

If conservation of angular momentum is a physical fact, how can it depend on my coordinate choice whether a coordinate is cyclic?

The conservation law never depends on the chart — only its visibility does. In polar coordinates the constant is the momentum conjugate to a coordinate that happens to be absent from $H$ , so it is found by inspection; in Cartesian coordinates the very same constant exists as the combination $xp_y-yp_x$ , but no Cartesian coordinate is cyclic, so inspection finds nothing and you would need Poisson-bracket or Noether machinery to detect it. "Cyclic" is a statement about the pair (system, chart). The practical moral: invest in coordinates aligned with the expected symmetry before building $H$ — and the theoretical moral, developed from the next article onward, is that transforming to charts with more cyclic coordinates is a solution method in itself.

Why exactly is the Lagrangian reduction incomplete if the momentum is conserved either way?

Conserved is not the same as eliminated. On the Lagrangian side, $\partial L/\partial q_k=0$ gives you the constant $p_k=\partial L/\partial\dot q_k$ — but this is an equation relating velocities, and $\dot q_k$ itself still appears inside $L$ , hence inside every remaining Euler–Lagrange equation. To use the constant you must solve for $\dot q_k$ and substitute — and substituting into $L$ directly is illegal (it changes the variational problem and yields wrong equations; the sign of the centrifugal term comes out backwards). The Hamiltonian never has this problem because velocities were already traded for momenta by the Legendre transformation: setting $p_k=\alpha$ in $H$ is a legitimate substitution of a constant, and the remaining canonical equations close among themselves. The Routhian is precisely the device that makes the elimination legal in Lagrangian dress.

Is the Routhian a Hamiltonian or a Lagrangian? The signs keep confusing me.

It is deliberately both, split by coordinate type: Hamiltonian in the cyclic directions (which it eliminates), Lagrangian in the live ones (which it keeps). The sign subtlety is that $R=\sum_{\text{cyc}}p_j\dot q_j-L$ has both $\partial R/\partial q_i$ and $\partial R/\partial\dot q_i$ equal to minus the corresponding derivatives of $L$ , so the Euler–Lagrange equation keeps its form with $R$ in place of $L$ — but the honest effective Lagrangian is $L_{\text{eff}}=-R$ . The place this bites is the central force: the azimuthal kinetic energy $+\ell^2/2mr^2$ in $L$ reappears in $L_{\text{eff}}$ as a potential term $-\ell^2/2mr^2$ subtracted, i.e. a repulsive barrier $+\ell^2/2mr^2$ inside $U_{\text{eff}}$ . If your steady-orbit radii come out wrong by a sign, this flip is almost always the culprit.

Does a cyclic coordinate still give a conserved momentum when H depends explicitly on time?

Yes. The momentum equation $\dot p_k=-\partial H/\partial q_k$ involves only the $q_k$ -dependence of $H$ ; explicit time dependence is irrelevant to it. So a bead on a wire being cranked with a prescribed time-dependent profile can perfectly well conserve some $p_k$ while its energy is being pumped. The two conservation questions — is $q_k$ absent (conserves $p_k$ ), is $t$ absent (conserves $H$ ) — are independent switches. This independence is also why cyclic-coordinate reduction remains useful for driven systems where energy methods fail.

How is this related to Noether's theorem — is a cyclic coordinate just a special case?

Yes, the simplest one. A coordinate absent from $H$ (equivalently $L$ ) means the dynamics is invariant under rigid shifts $q_k\to q_k+\epsilon$ ; the conserved Noether charge of that one-parameter symmetry is exactly $p_k$ . Noether's theorem generalizes in two directions cyclic coordinates cannot reach: symmetries that mix several coordinates (a rotation in Cartesian variables, where no single coordinate shifts), and symmetries acting on time itself. The Lagrangian-side account — including the theorem and its charge formula — lives in Cyclic Coordinates and Conservation of Conjugate Momenta and the Noether article; this page's contribution is the Hamiltonian mechanism ( $\dot p_k=-\partial H/\partial q_k$ ) and the reduction that the conserved charge finances.

Can every system be brought to a chart where all coordinates are cyclic?

No — and the boundary is one of the deepest facts of mechanics. Systems for which it is possible (given $n$ independent conserved quantities in involution) are called integrable in the Liouville sense: the central force, the heavy symmetric top, all small-oscillation problems. For them, Hamilton–Jacobi theory and action-angle variables construct the all-cyclic chart explicitly, and motion winds uniformly on invariant tori. Generic systems with $n\ge2$ degrees of freedom are not integrable — the three-body problem being the classic casualty — and no chart, however clever, makes all their coordinates cyclic; chaos lives in the obstruction. The dream of equation (5) is thus a precise diagnostic: how close a system can get to it measures how solvable it is.

The conserved p for a charged particle came out different from mv in my problem set. Did I make an error?

Probably not — you rediscovered that the canonical momentum, not the kinetic one, is what cyclicity conserves. For a particle in a magnetic field the Lagrangian contains $q\,\mathbf{v}\cdot\mathbf{A}$ , so the momentum conjugate to an ignorable azimuth is $p_\phi=mr^2\dot\phi+q\,rA_\phi$ — kinetic angular momentum plus a field contribution. It is this combined object that stays constant in an axisymmetric field (the basis of magnetic-mirror and tokamak confinement), while $mr^2\dot\phi$ alone drifts. The rule: always derive $p_k=\partial L/\partial\dot q_k$ honestly before freezing it; whenever velocity-dependent potentials are in play, conserved canonical momenta and $m\mathbf{v}$ part company.

When should I use the Routhian instead of just working with the reduced Hamiltonian?

Use the reduced Hamiltonian when you want phase-space tools: portraits in the $(r,p_r)$ plane, Liouville-volume arguments (see Liouville's Theorem), or a launchpad for canonical transformations. Use the Routhian when the remaining analysis is naturally Lagrangian — above all, small oscillations about steady motion. Routh's own use case is still the standard one: a steady state with uniformly advancing cyclic coordinates (a precessing top, an orbiting satellite, a spinning governor) appears in the Routhian description as an equilibrium of the live coordinates, so the entire small-oscillation toolkit (expand $L_{\text{eff}}$ to second order, read off frequencies) applies verbatim. The two routes always agree — Derivations B and C produce the same radial equation — so the choice is purely about which toolbox you want open.

Cyclic Coordinates and Conserved Momenta

Where this lives today

The physical picture

Cyclic coordinates in the Hamiltonian language

Which statement when

The Routhian: reduce like a Hamiltonian, solve like a Lagrangian

Reduction in practice: the central force, start to finish

The canonical-transformation dream

Limiting cases

Worked examples

1 · Satellite in Earth's gravity: freeze the angle, read the orbit (confidence builder)

2 · Spherical pendulum: one cyclic angle, one live angle (exam level)

Going further

Before this

Where this leads