Cyclic Coordinates and Conserved Momenta
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You should be comfortable with the Euler-Lagrange equations and the definition of conjugate momentum. Familiarity with Noether's theorem for continuous symmetries is helpful but not strictly required for the core mechanics here.
Why this matters
Cyclic coordinates provide the most direct mechanical link between symmetry and conservation.
In the previous lecture, we established that every continuous symmetry of the action yields a conserved quantity. While Noether's theorem is the general framework, the case of cyclic coordinates is the simplest and most frequently encountered instance. It allows us to reduce the order of the differential equations we must solve, effectively lowering the dimensionality of the problem. This technique is indispensable in celestial mechanics, atomic physics, and the analysis of rotating systems, where angular variables often decouple from radial ones.
The Idea in Plain Language
Imagine a bead sliding on a frictionless horizontal hoop. If the hoop is perfectly circular and uniform, the physics does not change if you rotate the entire system. The angle of the bead is a coordinate that describes its position, but the laws of motion do not depend on where the bead is on the hoop, only on how fast it is moving. In Lagrangian mechanics, if a coordinate does not appear in the expression for the Lagrangian, it is "cyclic." Because the system is indifferent to the value of this coordinate, the quantity associated with its rate of change—its momentum—must remain constant. This is a direct manifestation of rotational symmetry: because the system looks the same from every angle, the angular momentum is conserved.
Definition and Identification
A generalised coordinate \( q_k \) is termed cyclic (or ignorable) if it does not appear explicitly in the Lagrangian \( L(q, \dot{q}, t) \). Mathematically, this condition is expressed as:
It is crucial to distinguish this from the case where the coordinate appears in the kinetic energy term. For example, in polar coordinates, the radial coordinate \( r \) usually appears in the Lagrangian, but the angular coordinate \( \theta \) often does not, provided the potential energy is rotationally symmetric. If \( L \) depends on \( \dot{q}_k \) but not on \( q_k \), the coordinate is cyclic. Note that a coordinate can be cyclic in one coordinate system and non-cyclic in another. The property is not intrinsic to the physical degree of freedom but depends on the choice of generalised coordinates.
Conservation of Conjugate Momentum
The Euler-Lagrange equation for the coordinate \( q_k \) is given by:
If \( q_k \) is cyclic, the second term vanishes. We define the conjugate momentum \( p_k \) as:
Substituting this into the Euler-Lagrange equation yields:
This implies that \( p_k \) is a constant of motion. This result is a specific case of Noether's theorem, where the symmetry is a translation in the coordinate \( q_k \). The conservation of \( p_k \) allows us to replace the second-order differential equation for \( q_k \) with a first-order algebraic constraint, simplifying the integration of the equations of motion.
Reduction of Order
When a coordinate is cyclic, we can use the conservation of its conjugate momentum to reduce the complexity of the system. Suppose we have a system with \( N \) degrees of freedom, and \( m \) of them are cyclic. We can solve for the \( m \) cyclic coordinates directly from the conservation laws. The remaining \( N-m \) coordinates obey equations in which the constant momenta appear as parameters. Take care: eliminating the cyclic velocities by substituting directly into \( L \) gives the wrong sign for terms such as the centrifugal term; the correct reduced function is the Routhian \( R = L - \sum_k p_k \dot{q}_k \), or you can substitute into the equations of motion after deriving them, as in the worked example. This process is known as the reduction of order. It is particularly powerful in central force problems, where the angular coordinate is cyclic, allowing the angular momentum to be treated as a constant parameter in the radial equation.
The derivation
Cyclic Coordinates and Conserved Conjugate Momenta
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).
A common pitfall is assuming that if a coordinate is cyclic, its conjugate momentum is always the physical linear or angular momentum. This is true for standard Cartesian and polar coordinates, but in more complex generalised coordinates, \( p_k \) may be a combination of linear and angular momenta. Always compute \( p_k = \partial L / \partial \dot{q}_k \) explicitly rather than assuming a physical interpretation. C
Worked example
Problem. A particle of mass \( m \) moves in a plane under the influence of a central potential \( V(r) \). The Lagrangian in polar coordinates \( (r, \theta) \) is given by \( L = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - V(r) \). Identify the cyclic coordinate, find the conserved quantity, and write the equation of motion for \( r \).
- Identify the cyclic coordinate. The Lagrangian depends on \( r \) and \( \dot{r} \), but does not depend explicitly on \( \theta \). Thus, \( \theta \) is cyclic.
- Compute the conjugate momentum for \( \theta \): \( p_\theta = \frac{\partial L}{\partial \dot{\theta}} = m r^2 \dot{\theta} \). This is the angular momentum \( L_z \).
- Since \( \theta \) is cyclic, \( p_\theta \) is conserved. Let \( p_\theta = h \) (a constant).
- Use the conservation law to eliminate \( \dot{\theta} \): \( \dot{\theta} = \frac{h}{m r^2} \).
- Substitute this into the equation of motion for \( r \). The Euler-Lagrange equation for \( r \) is \( \frac{d}{dt}(m\dot{r}) - (m r \dot{\theta}^2 - V'(r)) = 0 \). Substituting \( \dot{\theta} \) gives \( m\ddot{r} - \frac{h^2}{m r^3} + V'(r) = 0 \): the radial motion is that of a one-dimensional particle in the effective potential \( V(r) + h^2/(2 m r^2) \).
Units check. \( m\ddot{r} \) has units \( \mathrm{kg \cdot m/s^2} = \mathrm{N} \). \( h^2/(m r^3) \) has units \( (\mathrm{kg \cdot m^2/s})^2 / (\mathrm{kg \cdot m^3}) = \mathrm{kg^2 \cdot m^4/s^2} / (\mathrm{kg \cdot m^3}) = \mathrm{kg \cdot m/s^2} = \mathrm{N} \). \( V'(r) \) is force, \( \mathrm{N} \). The units are consistent.
Three quick tests
- If a coordinate \( q \) is cyclic, what is the time derivative of its conjugate momentum?
Answer
Zero. \( dp/dt = 0 \). - Can a coordinate be cyclic in one coordinate system and non-cyclic in another?
Answer
Yes. Cyclicity depends on the explicit form of the Lagrangian in that specific coordinate system. - What is the physical interpretation of the conserved momentum for the angular coordinate in a central force field?
Answer
Angular momentum about the center of force.
Problem set for PU-201 →
Each derivation in this unit ends with five graded problems with full solutions.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.