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In Lagrangian formalism, when $\frac{\partial L}{\partial q} = 0$, the coordinate $q$ is called cyclic and a corresponding conserved quantity exists. But why is it called cyclic?

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That link may hint at the answer by analogy with action-angle variables - the angle variable moves along a cycle in phase space. I don't know that this is correct, though. – Chris White May 31 '13 at 22:43

In a finite-size system this situation usually corresponds to the rotational degree of freedom, that's why it is called "cyclic".

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Sorry for the -1, if you edit even a word I can take it away – anna v Jun 1 '13 at 6:52

At first I thought like @MaximUmanski that it referred to rotational degrees of freedom. I searched the index of Goldstein, "Classical Mechanics" second edition for the definition and found that the definitions are not clear, but that it does not refer to periodicity particularly.

In page 55 (par 2.6 . there exist free downloadable versions but one has to be at least on facebook ) where the first mention of cyclic coordinates is made he says

if the lagrangian of a system does not contain a given coordinate $q_j$ (although it may contain the corresponding velocity of $q_j$) then the coordinate is said to be cyclic or ignorable. The definition is not universal but is the customary one.

Then goes ahead with a long footnote quoting different definitions by different people and makes clear that

in addition "cyclic" is sometimes used in a different sense in connection with periodic variables.

In this lecture the usefulness of the cyclic identification is pointed out: it appears in the Hamiltonian formulation by reducing the number of differential equations for the problem at hand. I suspect that the term cyclic probably is coming from this function of the coordinates, in the sense that the equations can be solved with the ignorable (cyclic) coordinates giving a trivial contribution, the way periodic functions give a trivial contribution to the complexity of a solution.

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If only the cyclic coordinate $q(t)$ varies with time (if it doesn't, $q$ is superfluous), the Lagrangian, or the essential physical situation, doesn't vary. Hence the initial value of $q$ doesn't determine the path, which is only possible if the path is closed.

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protected by Qmechanic Jul 28 '14 at 15:53

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