physics2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Derivation

Cyclic Coordinates and Conserved Conjugate Momenta

D-129 Home PU-201 Threads symmetry · energy Depends on The Euler-Lagrange Equation, Noether's Theorem for Continuous Symmetries
Statement

If a generalised coordinate qk does not appear explicitly in the Lagrangian L(q, q̇, t) — that is, ∂L/∂qk = 0 — then its conjugate momentum pk = ∂L/∂q̇k is conserved along any physical trajectory: dpk/dt = 0. Such a coordinate is called cyclic (or ignorable).

Why it matters

This is the most direct and computationally useful conservation theorem in mechanics. Rather than solving the full set of coupled second-order equations, you inspect the Lagrangian by eye: every coordinate missing from it hands you a first integral of motion for free, reducing the order of the problem by one for each cyclic coordinate.

It is also the concrete, coordinate-level shadow of Noether's theorem. Absence of qk from L means L is invariant under the translation qk → qk + ε; the conserved pk is exactly the Noether charge of that one-parameter symmetry. Linear momentum, angular momentum, and (via time) energy all appear this way.

Assumptions
The dynamics follow from a Lagrangian obeying the Euler–Lagrange equations.If the system is not derivable from a variational principle (e.g. genuinely non-holonomic or dissipative forces outside L), the Euler–Lagrange equation used in Step 1 does not hold and the conclusion fails.
qk is a true coordinate and q̇k may appear in L.If q̇k were also absent then pk = ∂L/∂q̇k = 0 identically and the statement is empty — cyclicity is only informative when the velocity survives while the coordinate does not.
All non-conservative or constraint generalised forces have been folded into L or vanish.If a generalised force Qk(nc) acts on the cyclic coordinate, the correct equation reads dpk/dt = Qk(nc), so pk drifts rather than being conserved.
Derivation
1
d/dt ( ∂L/∂q̇k ) − ∂L/∂qk = 0
Euler–Lagrange equation for coordinate qk on the physical path (assumed prior result). A
2
pk ≡ ∂L/∂q̇k
Definition of the momentum conjugate to qk. A
3
dpk/dt − ∂L/∂qk = 0
Substitute the definition of Step 2 into Step 1. A
4
∂L/∂qk = 0
Hypothesis: qk is cyclic — it does not appear explicitly in L. A
5
dpk/dt = 0
Put Step 4 into Step 3; the driving term vanishes so the momentum has zero time-derivative. A
6
L(qk+ε, q̇k, t) = L(qk, q̇k, t) ⇒ Q = Σ (∂L/∂q̇j)(δqj/ε) = pk
Cross-check against Noether's theorem: with generator δqj = δjk ε the invariance ∂L/∂qk=0 gives conserved charge pk, confirming Step 5 independently. C
Result
∂L/∂qk = 0 ⟹ pk = ∂L/∂q̇k = constant

Reading. A coordinate that the Lagrangian ignores cannot exert any generalised force on its own momentum, so that momentum is a constant of the motion. Each cyclic coordinate supplies one first integral pk(q, q̇) = ck, lowering the effective order of the problem by one.

Units check. pk = ∂L/∂q̇k has units of [energy]/[q̇k] = [J·s]/[qk]. If qk is a length (m), pk is J·s/m = kg·m·s−1, ordinary linear momentum. If qk is an angle (dimensionless rad), pk is J·s = kg·m2·s−1, angular momentum. Both are consistent with a conserved quantity.

Limiting cases
  • Free particle, Cartesian x: L = ½m(ẋ²+ẏ²+ż²) has no x, so px = mẋ = const — conservation of linear momentum.
  • Central force, polar φ: L = ½m(ṙ²+r²φ̇²) − V(r) has no φ, so pφ = mr²φ̇ = const — conservation of angular momentum (Kepler's second law).
  • Uniform gravity, horizontal x: V = mgy depends only on y, so x is cyclic and horizontal momentum mẋ is conserved while vertical is not.
  • Axisymmetric top, precession angle ϕ: ϕ and spin ψ are both cyclic, giving two conserved momenta that make the top integrable.
Breaks when
  • The coordinate reappears through an explicit external field. If a potential V(qk, t) depends on qk — e.g. a charge in a spatially varying field along x — then ∂L/∂qk ≠ 0, so pk is no longer conserved.
  • Non-Lagrangian forces act. With friction, drag, or a non-holonomic constraint force on qk, the equation becomes dpk/dt = Qk(nc) ≠ 0; the momentum bleeds away even though qk is absent from L.
  • Cyclicity is coordinate-dependent. A quantity conserved in one chart need not be manifest in another. Choosing coordinates in which qk appears (e.g. Cartesian for a central-force problem) hides the conservation law even though the physics is unchanged — the theorem gives a sufficient, not necessary, test.
Failure modes
  • Confusing "absent coordinate" with "absent velocity". The theorem needs ∂L/∂qk=0, not ∂L/∂q̇k=0. Students who look for the missing velocity find nothing conserved.
  • Reading pk as m q̇k automatically. In curvilinear coordinates pφ = mr²φ̇, not mφ̇; you must differentiate the actual L.
  • Assuming energy conservation from a cyclic spatial coordinate. Cyclic qk conserves pk; energy conservation follows separately from ∂L/∂t=0, a distinct condition.
  • Ignoring velocity-dependent (magnetic) potentials. For a charge in a magnetic field, px = mẋ + qAx; forgetting the qA term gives the wrong conserved quantity (canonical vs kinetic momentum).
  • Fixing a coordinate by constraint and calling it cyclic. Holding qk constant removes a degree of freedom; it does not make pk a dynamical constant of motion.
Discussion

The result exposes the mechanical meaning of "momentum". The conjugate momentum pk is defined purely by how L depends on the velocity q̇k, and its conservation is controlled purely by how L depends on the coordinate qk. Position and velocity thus play complementary roles: the velocity dependence builds the momentum, the coordinate dependence drives it. When the drive is switched off, the momentum coasts.

Cyclic coordinates are the engine of the Routhian and Hamilton–Jacobi reduction schemes. Because pk is constant, one can pass to a reduced description (the Routhian) in which the cyclic coordinate is eliminated entirely and replaced by its conserved momentum as a parameter. A system with enough cyclic coordinates becomes trivially integrable — this is exactly why the symmetric top, the Kepler problem, and geodesics on surfaces of revolution all admit closed-form treatment.

At the deepest level the theorem is Noether's first theorem restricted to point (configuration-space) symmetries generated by a single coordinate translation. The conserved charge Q = Σj pj (∂qj/∂ε) collapses to pk when the generator is δqj = δjkε. In the Hamiltonian picture the same statement reads {pk, H} = −∂H/∂qk = 0, so a cyclic coordinate makes pk commute with the Hamiltonian — the classical precursor of a quantum operator that commutes with Ĥ and labels states by a good quantum number.

Common misconceptions. Cyclic does not mean "periodic"; it means "ignorable / absent from L". The name is historical. Also, conservation of pk does not imply qk itself is simple — qk(t) generally still varies in time; it is only its conjugate momentum that is fixed.

Worked examples
1
Planar Kepler orbit — conserved angular momentum. L = ½m(ṙ² + r²φ̇²) + GMm/r
Set up the Lagrangian in plane polar coordinates for a body of mass m about a fixed mass M. The angle φ is absent from L, so it is cyclic.
2
pφ = ∂L/∂φ̇ = m r² φ̇ = const ≡ ℓ
Differentiate L with respect to φ̇; cyclicity of φ makes this the conserved angular momentum.
3
m = 5.97×10²⁴ kg, r = 1.50×10¹¹ m, φ̇ = 1.99×10⁻⁷ rad·s⁻¹
Insert Earth's mass, orbital radius, and mean angular velocity (2π per year).
4
ℓ = (5.97×10²⁴)(1.50×10¹¹)²(1.99×10⁻⁷) = 2.67×10⁴⁰ kg·m²·s⁻¹
Evaluate: r² = 2.25×10²², times m gives 1.34×10⁴⁷, times φ̇.
ℓ = pφ ≈ 2.7×10⁴⁰ kg·m²·s⁻¹ (constant)

Reading. Earth's orbital angular momentum is fixed for all time by the cyclicity of φ; this is Kepler's second law (equal areas in equal times, since dA/dt = ℓ/2m). Units kg·m²·s⁻¹ confirm an angular momentum.

1
Charged particle in a uniform magnetic field B = B ẑ, gauge A = ½ B(−y, x, 0). L = ½m(ẋ²+ẏ²+ż²) + q(ẋAx+ẏAy+żAz)
Write the Lagrangian for a charge q in a magnetic field. The coordinate z is absent from both A and the kinetic term structure — it is cyclic.
2
pz = ∂L/∂ż = m ż + q Az = m ż = const
Differentiate with respect to ż; since Az=0 the canonical momentum equals the kinetic momentum along z, and cyclicity conserves it.
3
m = 9.11×10⁻³¹ kg (electron), ż = v∥ = 3.0×10⁶ m·s⁻¹
Take an electron with a fixed velocity component along the field axis.
4
pz = (9.11×10⁻³¹)(3.0×10⁶) = 2.73×10⁻²⁴ kg·m·s⁻¹
Multiply. Because z is cyclic this value does not change as the electron spirals.
pz = m v∥ ≈ 2.7×10⁻²⁴ kg·m·s⁻¹ (constant)

Reading. A magnetic field does no work and cannot change the velocity component along itself; cyclicity of z makes this exact. The particle executes uniform drift along z superposed on circular gyration in the x–y plane. Units kg·m·s⁻¹ confirm a linear momentum.

Problems
  1. A bead of mass m slides on a frictionless horizontal wire lying along x, with L = ½mẋ². State the cyclic coordinate and its conserved momentum, and evaluate for m = 0.20 kg, ẋ = 3.0 m·s⁻¹.
    Solution x is cyclic since ∂L/∂x = 0. Conjugate momentum px = ∂L/∂ẋ = mẋ = (0.20)(3.0) = 0.60 kg·m·s⁻¹, conserved. No horizontal force acts, so momentum stays constant.
  2. For a particle in a central potential, L = ½m(ṙ² + r²φ̇²) − V(r). Show pφ is conserved and compute it for m = 2.0 kg, r = 0.50 m, φ̇ = 4.0 rad·s⁻¹.
    Solution φ does not appear in L (only φ̇ and r do), so ∂L/∂φ = 0 ⇒ pφ = mr²φ̇ = const. Numerically pφ = (2.0)(0.50)²(4.0) = (2.0)(0.25)(4.0) = 2.0 kg·m²·s⁻¹.
  3. A projectile moves under gravity with L = ½m(ẋ² + ẏ²) − mgy. Identify which momentum is conserved and which is not, and give dpy/dt for m = 1.5 kg, g = 9.81 m·s⁻².
    Solution x is cyclic ⇒ px = mẋ conserved. y is not cyclic since ∂L/∂y = −mg ≠ 0. Euler–Lagrange: dpy/dt = ∂L/∂y = −mg = −(1.5)(9.81) = −14.7 N. So vertical momentum decreases at 14.7 kg·m·s⁻² while horizontal is constant.
  4. A symmetric top has L = ½I1(θ̇² + φ̇²sin²θ) + ½I3(ψ̇ + φ̇cosθ)² − Mgℓcosθ. Which Euler angles are cyclic, and what are the corresponding conserved momenta (symbolically)?
    Solution φ and ψ are absent from L (only their velocities and θ appear), so both are cyclic. pψ = ∂L/∂ψ̇ = I3(ψ̇ + φ̇cosθ) = const (spin angular momentum about the symmetry axis). pφ = ∂L/∂φ̇ = I1φ̇sin²θ + I3(ψ̇+φ̇cosθ)cosθ = const (angular momentum about the vertical). θ is not cyclic because Mgℓcosθ and the sin²θ, cosθ terms carry it.
  5. An electron drifts along a magnetic field B = 0.50 T with parallel speed v∥ = 1.0×10⁶ m·s⁻¹, gauge chosen so Az = 0 and z is cyclic. Find the conserved pz, and state whether the perpendicular canonical momentum px is also conserved in this gauge.
    Solution z cyclic ⇒ pz = mż + qAz = mv∥ = (9.11×10⁻³¹)(1.0×10⁶) = 9.1×10⁻²⁵ kg·m·s⁻¹, conserved. With the symmetric gauge A = ½B(−y,x,0), Ax = −½By depends on y so x is cyclic and px = mẋ − ½qBy is also conserved (canonical, not kinetic). Note the value of the conserved px is gauge-dependent, unlike pz here where Az=0.