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Here is what I mean:

In Lagrangian mechanics, we have the equation $$ \frac{\mathrm{d}}{\mathrm d t} \frac{\partial L}{\partial \dot q_i} = \frac{\partial L}{\partial q_i},\quad i = 1,2, \cdots, n. $$ and, a cyclic coordinate is the one that is not explicitly included in $L$: e.g., if $q_n$ is a cyclic coordinate, then $$ L = (q_1, \cdots, q_{n-1},\dot q_1, \cdots, \dot q_{n-1},\dot q_n), \quad \text{and thus}\quad \frac{\partial L}{\partial q_n} = 0. $$

And what I am curious about is: whether there is a Lagrangian system that essentially has no cyclic coordinates.

"Essentially" means that for any general coordinate system we use, the expression of $L$ contains no cyclic coordinate. This excludes the case of, e.g. the case of Kepler's two body problem, which has cyclic coordinate in Cartesian coordinate system, but have cyclic coordinates in polar coordinate.

If the answer of this question is YES, then furthermore, I am curious what if we limit to the "local" case: we do not search for the coordinate that can cover the entire configuration space, but instead, only consider a local coordinate around a general point, so is there any Lagrangian system that essentially has no cyclic coordinates even we take the local coordinates into consideration?

Editted on Dec 15 2024:

Thanks for @Eli 's reminding. I realized that I need to clarify this:

I assume that $L$ is autonomous (time independent), i.e. it is $L(q, \dot q)$ not $L(q, \dot q, t)$, so that time evolution is a symmetry, and energy $E_L = \sum_{i} \dot q_i \frac{\partial L}{\partial \dot q_i} - L$ is conserved. I want to know whether there is some $L$ that has no other continuous symmetries other than this one.

Mr. Egg
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2 Answers2

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For technical reasons$^1$, let us pose OP's question for Hamiltonian (instead of Lagrangian) systems, i.e. if there always exist a coordinate system in the phase space $M$ so that all position coordinates are cyclic (wrt. to the Hamiltonian $H$)?

  1. Given a fixed point $x_{(0)}\in M$ in phase space, under mild regularity assumptions, there always exists locally (in a sufficiently small open Darboux$^2$ neighborhood of $x_{(0)}$) an $n$-parameter complete solution for Hamilton's principal function $$S(q^1, \ldots, q^n; P_1, \ldots,P_n; t)$$ to the Hamilton-Jacobi equation, where $$P_i, \qquad i\in\{1, \ldots, n\},$$
    are integration constants. Recall that $S$ is a generating function for a canonical transformation (CT) of type 2. In the new coordinates $(Q^1, \ldots, Q^n;P_1, \ldots, P_n)$, the Kamiltonian $K\equiv 0$ is identically zero, so there all the position coordinates $(Q^1, \ldots, Q^n)$ are cyclic (wrt. to the Kamiltonian $K$), cf. OP's question.

  2. A more interesting question is if there exist globally defined cyclic variables or not? This is intimately linked to integrability, or lack thereof, cf. e.g. this related Phys.SE post.

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$^1$ For the Legendre transformation between the Lagrangian and Hamiltonian formalisms, see e.g. this Phys.SE post.

$^2$ A Darboux neighborhood here means a neighborhood where there exists a set of canonical coordinates aka. Darboux coordinates $(q^1, \ldots, q^n;p_1, \ldots, p_n)$, cf. Darboux' Theorem.

Qmechanic
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A time independent Lagrangian system has time as cyclic coordinate and its corresponding dynamic variable, the energy, is a constant.

The general fact is that any cyclic coordinate corresponds to a conserved quantity in the Poisson algebra.

So the question reduces to: Are there canonical, time independent system of equations motion $$\dot p = -\nabla_q H(p,q), \quad \dot q = \nabla_p H(p,q)$$ that cannot transformed to a Hamiltoniam independent of $q_1:$ $$ H(p_1, \ p_2\dots,\quad q_2, \dots)$$ via a general canonical transformation.

So the answer is no: Any (locally Lipshitz-continuous) n-dimensional canonical system has a (local) solution (in the neighbourhood of the start point in phase space) by a canonical transformation to a set of n constant generalized momentum variables and n cyclic coordinates.

Is this an miracle? No, simply reverse the solution, expressing the start values by the values at later time t (as long as they are partially or completely invertible)

$$p(t)= P(p(t_0),q(t_0),t) \quad q(t) = Q(p(t_0),q(t_0),t)$$

to

$$p(t_0)=PQ^{-1}(p(t),q(t),t)_p \quad q(t_0) = PQ^{-1}(p(t),q(t),t)_q$$

By definition all these dynamic variables are independent of t. The difficulty is, that symmetry groups of canonical momentum sets with more than 2 generators are complicated beasts, so that the 3-body problem already does not allow for an algebraic solution. An algebraic representation of its hidden constants of motion in terms of the cartesian coordinates and their euclidean momentums defining the free Lagrangian in general does not exist. Numerically no problem, except for error explosion in time.

Numerically then the search for nearly constant momentum variables and nearly cyclic coordinates is the only way to produce meaningful approximations for macroscopically long times. In astronomy one tries to control the longtime evolution of nearly conserved quantities, eg the slow movement of the perihel or the precession of the angular momentum axis.

So the search for cyclic coordinates and conserved momentums is meaningful: Anything goes, and many solutions of special systems hav been found by very obscure procedures that try to find a canonical map to a partially solvable system.

Roland F
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