# Conservation of angular momentum using symmetry properties

Goldstein pg 59

It can be shown that if a cyclic coordinate $$q_{j}$$ is such that $$d q_{j}$$ corresponds to a rotation of the system of particles around some axis, then the conservation of its conjugate momentum corresponds to conservation of an angular momentum.

Suppose we have a two-body central force problem and reduce it to a single body problem. Then can we use the above quoted text to prove that the angular momentum is conserved?

While the quoted text above is correct and gives a worded explanation on how angular momentum is conserved in a two-body central force problem, generally, more rigor is required for a proof. We know that given the Lagrangian of the system the generalized momentum is defined as: $$p_j = \frac{\partial{L}}{\partial{\dot{q_j}}}$$ and also the Lagrange equations of motion are expressed by: $$\dot{p}_j=\frac{\partial{L}}{\partial{q_j}}$$ The Lagrangian for a two-body system (reduced to a one-body problem) of reduced mass $$\mu$$ moving in a central-force field described by the potential function $$U(r)$$ in polar coordinates is: $$L=\frac{1}{2}\mu(\dot{r}^2+r^2\dot{\theta}^2)-U(r)$$ from here we can take our angular momentum conjugate to $$\theta$$ as $$\dot{p}_\theta = \frac{\partial L}{\partial \theta}=0=\frac{d}{dt}\frac{\partial{L}}{\partial{\dot{\theta}}}$$ since the Lagrange equation of motion in $$\theta$$ is $$\frac{\partial L}{\partial \theta}-\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}=0$$ therefore $$\frac{d}{dt}\frac{\partial L}{\partial \dot{\theta}}=\frac{d}{dt}(p_\theta)=\texttt{constant}$$ Hence, angular momentum is conserved. And this all relies on the fact that $$\theta$$ is a cyclic coordinate which made its conjugate momentum zero as the quoted text states.