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

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 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?

• You are mixing polar and Cartesian coordinates. Work in either one but do not mix. – my2cts Mar 15 at 7:32
• Hint: You need to use rotational symmetry of potential. – Qmechanic Mar 15 at 7:55
• In $- m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k$ we have $\partial ^i \phi(x) = \frac{\partial \phi}{\partial x} \partial ^i x$, where $x = \sqrt{x^a x^a}$, a= 1,2,3, hence $- m \partial ^i \phi(x)\epsilon_{ijk}x^j \theta^k = - m \frac{\partial \phi}{\partial x} \frac{x^i}{x} \epsilon_{ijk} x^j \theta^k = 0$, because $\epsilon_{ijk} x^i x^j = 0$. Is this correct? – Lil'Gravity Mar 15 at 19:12

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.