Cyclic coordinates and Routh reduction

A coordinate is cyclic, or ignorable, when the partial derivative of the Lagrangian with respect to that coordinate vanishes. Its Euler–Lagrange equation then says the conjugate momentum is constant. The momentum can depend on the other coordinates and velocities, so a constant momentum need not imply a constant cyclic velocity. In a central planar potential, the polar angle is cyclic and angular momentum is fixed, but the angular speed changes whenever the radius changes.

Reduction fixes a value of this conserved momentum and solves its defining equation for the cyclic velocity. This requires local invertibility in the eliminated velocity direction. If the relevant velocity Hessian is singular, the simple procedure can fail or require additional constraints. State the momentum value and the domain where inversion works. For a central particle with nonzero radius and positive mass, the angular momentum relation can be inverted directly; the polar origin remains a coordinate singularity.

Use the convention R=L−pφ φdot, evaluated after solving for φdot at fixed pφ=ℓ. Other books define the negative Routhian, so the convention and corresponding equations must travel together. With our convention, the remaining coordinates obey ordinary Euler–Lagrange equations for R at fixed ℓ. The subtraction cancels the unwanted chain-rule terms caused by the eliminated velocity's dependence on the remaining coordinates. It is a partial Legendre transform, leaving the other velocities as velocities.

For central motion, this produces a reduced Lagrangian with an effective potential equal to the original potential plus a positive centrifugal barrier. Direct substitution into L instead leaves a positive angular kinetic term inside a would-be reduced Lagrangian. Varying that expression at fixed ℓ gives the wrong sign for the centrifugal contribution. The error is not repaired by calling angular kinetic energy a potential afterward. The transform defines which variables are held fixed during the variation.

Once the radial motion is known, integrate φdot=ℓ/(mr²) to reconstruct the angle, adding an initial angle. Reduction therefore removes an equation from the coupled solve but does not erase the physical orientation. It also selects a momentum sector: comparing equilibria at different ℓ compares different reduced problems. If an external motor instead maintains a prescribed angular speed, it exchanges angular momentum with the system. That driven problem cannot be analyzed by pretending the same fixed-ℓ effective potential applies.

Use the quadratic-function lab to inspect the positive-curvature bare harmonic potential in the example. Predict how doubling its coefficient changes the tangent slope at the same positive position, then test it. Its controls show only y=ax², its tangent, and its integral. They do not include the inverse-square centrifugal barrier, eliminate a coordinate, or reconstruct an orbit. The missing barrier must be included on paper when finding the actual circular radius and radial force. This deliberate limitation helps separate a potential ingredient from the complete reduced model.