Conservation of Angular Momentum when potential energy is function of position I have read that the angular momentum $(L)$ of a body is conserved if its potential energy is a function of solely its position vector. For example, the motion of planets on their orbits. I have two questions regarding this concept:

*

*How do we prove this?

*Is the converse also true?

 A: Angular momentum is conserved if your system is invariant under any arbitrary rotation.
To prove that angular momentum is conserved, note that if you calculate the force from the potential, the force itself will be parallel to the position vector.
\begin{equation}
\mathbf{F}(\mathbf{r}) = -\mathbf{\nabla} \Phi(r) =- \frac{d \Phi(r)}{dr} \frac{dr}{d \mathbf{r}},
\end{equation}
where only the chain-rule was used to find the derivative. The first term in the product does not depend on $\mathbf{r}$, only on $r$ (it depends only on the magnitude of $\mathbf{r}$). The second term is the radial unit vector $\mathbf{\hat{r}}$. Now note, that the force has a magnitude and a direction. The magnitude is contained in the first term, the direction is contained in the second term. If you calculate the torque that this field exerts on an object, you find that it is zero:
\begin{equation}
\mathbf{M} = \mathbf{r} \times \mathbf{F} = -  \mathbf{r} \times \left(\frac{d \Phi(r)}{dr} \frac{dr}{d \mathbf{r}}\right) = -\frac{d \Phi(r)}{dr} \cdot \mathbf{r} \times\mathbf{\hat{r}} = 0
\end{equation}
because $\mathbf{r} \times \mathbf{\hat{r}}$ = 0 (property of cross product).
So $\Phi (r) \rightarrow \mathbf{M}=0$ thus angular momentum is conserved.
I think that there might be some special potentials for which the angular momentum is conserved but the potential itself does not depend only on the magnitude of the position vector. This is why I would say that the converse might not be true.
A: The conservation of angular momentum is due to the symmetry of some physical systems under rotation, and is a special case of Noether's theorem.
To understand this, consider the Euler-Lagrange equation for a system:
$$\frac{\partial L}{\partial q_i} - \frac{d}{dt} \frac{\partial L}{\partial \dot{q}_i} = 0$$
Where $q_i$ and $\dot{q}_i$ are the generalized positions and velocities of particles in the system, and $L \left( q_i, \dot{q}_i, t \right)$ is the Lagrangian of the system, which equals the difference of total kinetic and potential energies in Newtonian mechanics:
$$\displaystyle{ L \left( q_i, \dot{q}_i, t \right) = \sum_i T \left( q_i \right) - V \left( \dot{q}_i \right) }$$
Now, the quantity $\frac{\partial L}{\partial \dot{q}_i}$ can be thought of as the conjugate momentum, $p_i$ , of position $q_i$, so we may rewrite the Euler-Lagrange equation as,
$$\frac{\partial L}{\partial q_i} = \frac{dp_i}{dt}$$
Obviously, if the Lagrangian $L$ is not a function of some coordinate $q_i$, the corresponding conjugate momentum is conserved,
$\frac{dp_i}{dt} = \frac{\partial L}{\partial q_i} = 0$
Such coordinates are also known as 'cyclic' coordinates. Thus, if a system is symmetric under rotations, all angle parameters become cyclic coordinates, and the corresponding conjugate momenta are conserved. These momenta happen to be the components of the angular momentum vector.
For example, consider the motion of one particle, characterized by two coordinates which we shall take as polar coordinates $\left( r, \theta \right)$. A system with rotational symmetry then has a Lagrangian of the form,
$L = \frac{1}{2} mv^2 - V \left( r \right) \\ L = \frac{1}{2} m r^2 \dot{\theta}^2 - V \left( r \right)$
Thus, the conjugate momentum of $\theta$ is,
$p_\theta = \frac{\partial L}{\partial \dot{\theta}} = mr^2 \dot{\theta} = mvr$
And $\frac{d p_\theta}{dt} = \frac{\partial L}{\partial \theta} = 0$
Voila, this conserved quantity is nothing but angular momentum. Similarly, symmetry under translation in space gives rise to conservation of linear momentum; translation in time implies conservation of energy; and so on.
In Newtonian mechanics, the kinetic energy term in the Lagrangian only depends on generalized velocities, so only the potential energy term needs to be independent of some coordinate for conservation of its conjugate momentum.
However, in general, it is not always possible to separate the Lagrangian into kinetic and potential energies, so we must sure that each term is not a function of generalized position.
