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Say I am working on a problem where I have two generalized coordinates $q_1,q_2$ and my Lagrangian is independent of $q_1$, i.e. I have $L(\dot{q}_1,q_2,\dot{q}_2)$. Then we say that $q_1$ is a cyclic coordinate, and that the corresponding momentum $\frac{\partial L}{\partial \dot{q_1}}$ is constant.

But if I compute $\frac{\partial L}{\partial \dot{q}_1}$ in my case, it is not constant, i.e. if I take $\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_1}$ it is not equal to $0$? What's going on?

Example, but I really just want to understand conceptually: Spherical pendulum with arm length $a$ $$L=\frac12 m(a^2\dot{\theta}^2+a^2\sin^2(\theta)\dot{\phi}^2)-mga(1-\cos(\theta))$$ since $\phi$ doesn't appear, we say that it is cyclic and hence corresponding generalised momentum is constant: $$p_\phi = ma^2\sin^2(\theta)\dot{\phi}$$ but $$\dot{p_\phi}=2ma^2\cos(\theta)\dot{\theta}+ma^2\sin^2(\theta)\ddot{\phi}\ne0.$$

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    $\begingroup$ We would probably need to see the Lagrangian in question and what you have done to be much help. $\endgroup$
    – Luke Pritchett
    Oct 17, 2017 at 15:34
  • $\begingroup$ @LukePritchett Wait are you saying what I suggest is not possible? I just meant conceptually. Apparently if $q_1$ doesn't appear we say that the corresponding momentum is constant, and I would think this means that $\dot{p}_1=0$, but it doesn't need to it seems? $\endgroup$
    – Terry E
    Oct 17, 2017 at 15:43
  • $\begingroup$ How do you conclude the very last inequality (v3)? $\endgroup$
    – Qmechanic ♦
    Oct 17, 2017 at 15:52
  • $\begingroup$ @Qmechanic Well, the LHS doesn't vanish identically firstly, so I suppose I am meant to conclude some physical phenomenon causes this to vanish? $\endgroup$
    – Terry E
    Oct 17, 2017 at 15:54

2 Answers 2

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The equations $\frac{\partial L}{\partial q_i}=\frac{d}{dt}\frac{\partial L}{\partial \dot q}$ define the equations of motion. They do not create quantities that are identically equal $\frac{\partial L}{\partial q_i}\equiv\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}$.

In your example, $\frac{\partial L}{\partial \phi}=0$, so it creates an equation $\dot p_\phi=0$ which is not an identity $\dot p_\phi\equiv 0$. The resulting equation $$2ma^2\cos(\theta)\dot\theta+ma^2\sin^2(\theta)\ddot\phi=0$$ gives you an equation of motion. Solving for $\ddot\phi$ yields $$\ddot\phi=2\cot(\theta)\csc(\theta)\dot\theta$$

This makes sense physically. If $\theta$ isn't changing, then it's following a circular path in the horizontal plane at constant speed. $\dot\phi$ will also be rapidly changing if it's following a path near $\theta=0$ or $\theta=\pi$ and at a minimum when at $\theta=\frac{\pi}{2}$.

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OP's question seems resolved by the fact that a constant of motion is a constant on-shell, i.e. modulo the equations of motion. It is not necessarily a constant off-shell, i.e. it is not necessarily identically equal to a constant.

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  • $\begingroup$ Does this mean when I take the E-L equation for $q_1$ (or in example $\phi$) I won't get nothing out of it, but rather some valuable piece of information, for using when I obtain my equations of motion? I mean that, in the example, after finding that $\phi$ is a cyclic coordinate, would I then only calculate the E-L equation for $\theta$? $\endgroup$
    – Terry E
    Oct 17, 2017 at 15:59
  • $\begingroup$ EL eq. for $\phi$ yields a constant of motion. $\endgroup$
    – Qmechanic ♦
    Oct 17, 2017 at 16:09

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