I was studying the Goldstein Classical Mechanics books and I became confused of this definition here:

Conservation Theorem: If the Generalized Coord $q_j$ is cyclic or ignorable, the corresponding Generalized (or Conjugate) Momentum, pj is conserved.

Well it seems to be a simple definition of a Conservation Theorem by cyclic coordinate. But I was wondering about cases when some coordinates have constrictions.

If a coordinate qj has a holonomic constriction by g(qj)=0, we can say that even if ∂L/∂qj=0, which states that L is not explicit of qj, the generalized momentum is not conserved due to Lagrangian multiplicator terms.

Now the question is, for a coordinate to be cyclic, does it only have to be not explicit about the Lagrangian or does the corresponding Generalized Momentum has to be conserved also?

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Which one is correct among the two images?

Here is an example about the constraint problem: enter image description here

enter image description here

enter image description here

If the rolling body has a constraint of no-slip condition (x=Rtheta), friction is applied on the x and theta coordinate. Then we can say the generalized coordinate of x and theta is not conserved. theta is not explicit about L, but can we say theta is cyclic?

Well, as the system has a holonomic constraint(no-slip) the system has no coordinates which the corresponding generalized momentum is conserved. Therefore I was wondering:

enter image description here even if the generalized momentum is not conserved, can the coordinate be cyclic? What if we, suppose change the Lagrangian in a form like: enter image description here It's basically the same Lagrangian except the holonomic constaint is applied. Can we still say the coordinate theta be cyclic?

  • $\begingroup$ I mean, it is a definition, if L does not depend explicitly of X, then X is cyclic. I am not understanding your doubt. The conservation of X momentum comes if, being the lagrangian T-V, the potential not being dependent of $\dot X$, but this is just another thing. $\endgroup$
    – LSS
    May 11, 2021 at 2:26
  • 1
    $\begingroup$ Please write the question in Mathjax $\endgroup$
    – Physiker
    May 11, 2021 at 7:12

2 Answers 2

  1. Definition. A coordinate $q^j$ is cyclic if the Lagrangian $L$ doesn't depend on it: $\frac{\partial L}{\partial q^j}=0$.

  2. The corresponding momentum $p_j=\frac{\partial L}{\partial \dot{q}^j}$ is not necessarily conserved unless the Lagrange equations is of the form of the Euler-Lagrange equations, cf. e.g. my Phys.SE answer here.


Lets make it clear

Case I

you have two degrees of freedom $~x~,\theta$

the Lagrange function is:

$$\mathcal{L}=\frac{m}{2}\,\dot{x}^2+\frac{I}{2}\dot{\theta}^2+m\,g\,\sin(\phi)\,x-(x-r\,\theta)\,\lambda\tag {1}$$

the last term is added because you are not working with generalized coordinate, hence you have to add the rolling condition $~x-r\,\theta~$ and the generalized constraint force $\lambda$


$$\frac{\partial\mathcal{L}}{\partial \theta}=r\,\lambda\ne 0$$ $$\frac{\partial\mathcal{L}}{\partial x}=m\,g\,\sin(\phi)-\lambda\ne 0$$

Case II

from the rolling condition you can obtain $~x=r\,\theta~,\Rightarrow\,\dot x=r\,\dot\theta$ so $~\theta$ is now generalized coordinate.

from equation (1)

$$\mathcal{L}=\frac 12\,m{{r\,\dot\theta}}^{2}+\frac 12\,I{\,\dot\theta }^{2}+mg\sin \left( \phi \right) r \theta $$


$$\frac{\partial\mathcal{L}}{\partial \theta}=m\,g\,\sin(\phi)\,r\ne 0$$

thus either $x$ nor $\theta$ are cyclic coordinates

  • $\begingroup$ @dc07650 your last Lagrange function is wrong $\endgroup$
    – Eli
    May 11, 2021 at 12:53

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