# How do you know if a coordinate is cyclic if its generalized velocity is not present in the Lagrangian?

Goldstein's Classical Mechanics says that a cyclic coordinate is one that doesn't appear in the Lagrangian of the system, even though its generalized velocity may appear in it (emphasis mine). For example, in the the equation $$L=\frac{1}{2}m\bigl[\dot r^2+r\dot \theta^2+r^2\dot \phi^2\sin^2\theta\bigr]\tag1$$ $\phi$ is the cyclic coordinate.

My question is: If the last term in equation (1) were missing, would $\phi$ still be a cyclic coordinate, and how would we know? What I'm asking is, how would we identify that a coordinate is cyclic if its generalized velocity was not present to give us a hint?

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You can consider whatever coordinates are describing your system. The Lagrangian you wrote (which I'm assuming is supposed to have two powers of $r$ in the second term) is for a free particle in spherical coordinates, and so you probably knew before writing it down that you were dealing with $(r, \theta, \phi)$.
Of course, if you ever write down a Lagrangian and neither the coordinate nor its time derivative appear, it probably means you introduced an unnecessary coordinate. For instance, suppose you had $$L = \frac{1}{2} m \Big(\dot{r}^2 + r^2 \dot{\theta}^2\Big),$$ as you suggested. Then the Euler-Lagrange equation will simply be $0 = 0$ for anything other than $r$ or $\theta$. Nothing interesting is happening to any other coordinates, which we can formally call cyclic, even if the kinetic and potential energies might independently rely on them.