# Spherical Potential and Angular Momentum Conservation

I have always found it clear that since a spherical potential has all components of angular momentum conserved since the entire system is symmetric under any rotation. However, I was trying to prove this to myself using Noether's theorem. The Lagrangian in spherical coordinates for such a system:

$$L = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta^2} + r^2 \sin^2(\theta) \dot{\phi}^2) - V(r).$$

Since the lagrangian does not have an explicity $$\phi$$ dependence, the conservation holds. However, when I try this on the $$\theta$$ axis, I am running into some troubles. Let $$\theta \rightarrow \theta + \lambda$$. Then by Noether's theorem, I need to show that the lagrangian does not change in the first order of $$\lambda$$:

$$L' = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta^2} + r^2 \sin^2(\theta + \lambda) \dot{\phi}^2) - V(r).$$

The problem is with the sine term. My first intuition is to taylor expand:

$$\sin(x) = \sin(\theta) + \cos(\theta)(x - \theta) - \frac{1}{2} \sin(\theta) (x - \theta)^2+...$$

$$\sin(\theta + \lambda) = \sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...$$

$$\sin^2(\theta + \lambda) = (\sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...)(\sin(\theta) + \cos(\theta)\lambda - \frac{1}{2} \sin(\theta) \lambda^2+...)$$ $$= \sin^2(\theta) + 2\sin(\theta)\cos(\theta) \lambda + ....$$

Which is not a first order change. In other words:

$$\frac{\partial L}{\partial \lambda} = \frac{1}{2}mr^22\sin(\theta)\cos(\theta)\dot{\phi^2} \neq0.$$

However, I know intuitively that around $$\theta$$, there is symmetry. What is exactly wrong in my line of logic?

The issue is that the transformation $$\theta\to\theta+\delta\theta$$ is not a rotation. This explains why the Lagrangian is not conserved ad does not give you an integral of motion. Remember the definition of $$\theta$$ is spherical coordinates, it is not periodic like $$\phi$$ so your transformation does not even make much sense.
Polar coordinates are more useful in the Hamiltonian formalism. Using separation of variables, you can easily get the conservation of $$L^2$$ without noticing the conservation of the other components of angular momentum. The corresponding Hamiltonian is: $$H=\frac{1}{2m}\left(p_r^2+\frac{p_\theta^2}{r^2}+\frac{p_\phi^2}{r^2\sin^2\theta}\right)+\frac{K}{r^n}$$ The component $$L_z=p_\phi$$ is conserved corresponding to rotations about $$z$$ and you also get the conservation of $$L^2=p_\theta^2+\frac{p_\phi^2}{\sin^2\theta}$$ as announced.