# Cyclic integral?

Can anyone explain me what is cyclic integral and give me some instructions how to show that there exist cyclic integral for Lagrangian
$$L~=~\frac{1}{2}ma^2{\dot {\theta}}^2+\frac{1}{2}ma^2{\dot {\phi}}^2\sin^2\theta+mga\cos\theta?$$

Given a Lagrangian $L(q^1, \dots, q^N, \dot q^1, \dots, \dot q^N)$ defined on coordinates $q_1, \dots q_N$ and their corresponding velocities $\dot q_1, \dots, \dot q_N$, a cyclic coordinate $q^{i}$ is one on which the Lagrangian doesn't explicitly depend; $$\frac{\partial L}{\partial q^{i}} = 0$$ Since the Euler Lagrange equations for each coordinate $q^k$ is $$\frac{d}{dt}\frac{\partial L}{\partial \dot q^k} = \frac{\partial L}{\partial q^k}$$ The Euler Lagrange equation for the cyclic coordinate $q^i$ has vanishing right hand side and becomes $$\frac{d}{dt}\frac{\partial L}{\partial \dot q^i} = 0$$ so that the quantity (incidentally called the canonical momentum corresponding to the coordinate $q^i$) $$p_i=\frac{\partial L}{\partial \dot q^i}$$ is conserved. Such a conserved quantity is often referred to as a first integral, and although I've never personally seen the term "cyclic integral," I'm guessing that it is simply the conserved quantity generated from the existence of a cyclic coordinate as described above.
In the Lagrangian you have written down, the coordinate $\phi$ is cyclic since it doesn't appear explicitly in the Lagrangian. As a result the quantity $\partial L/\partial \dot\phi$ is conserved, and this is probably the "cyclic integral" you are looking for.
• The term "cyclic integral" as an alternative to the usual "integral of motion" turns out to be the more natural of the two in Hamiltonian mechanics, where the symplectic integral over an orbit is zero: $\oint \space dp\land dq =0$. Jun 30, 2013 at 23:00