Lagrangian of a particle in a gravitational potential and conservation of angular momentum

Question

I'm trying to prove via Noether's Theorem that the angular momentum of a massive particle in a gravitational field is conserved. The attempt follows:

OBS: I'm working in the #D Euclidean space so I'll make no distinction between upper and lower indices

The lagrangian of this particle is given by: $$L = \frac{1}{2} m \dot{x}^i \dot{x}_i - m \phi(x)$$ where $\phi(x)$ is the gravitational potential.

For an infinitesimal rotation by an infinitesimal angle vector $\vec{\theta}$ with components $\theta ^k$ such that $x^i \rightarrow x^i + \epsilon_{ijk}x^j \theta ^k$, we have

\begin{equation} \delta L = - m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k + m\dot{x}^i \epsilon_{ijk} \dot{x}^j \theta ^k. \end{equation}

The second term in RHS is zero because $\dot{x}^i \dot{x}^j$ is symmetric and $\epsilon_{ijk}\theta ^k$ is antisymmetric.

My problem comes with the first term in RHS, because I think that it should be zero, such that I would get the conserved quantity $$Q = \epsilon_{ijk}m\dot{x}^i x^j \theta^k = \vec{l} \cdot \vec{\theta},$$ with $\vec{l}$ being the angular momentum vector, and $\vec{l} \cdot{\vec\theta}$, the angular momentum in the direction of $\theta$. It is necessary that $- m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k = 0$ to get conservation of angular momentum or did I make some mistake?

Answer 1

You have an extra time derivative in your expression for $Q$, there should be just one. (Your present expression for $Q$ vanishes, for the same reason you have described in the question). Having made that alteration, if we demand that $\frac{dQ}{dt}=0$, we will need to substitute the equation of motion. Upon substitution, this turns out to be precisely the first term.