# Without Lagrangian, show that angular momentum is conserved

• jack476
In summary, the conversation discusses the possibility of showing that rotational invariance about an axis implies the conservation of angular momentum about that axis without using the Lagrangian formalism or Noether's theorem. The proposed solution involves considering the interactions of a system with the outside world in terms of a potential energy function and using vector geometry to show that the force and torque experienced by the system are both 0, leading to the conservation of angular momentum. The person asking for help is unsure if this solution is valid and is seeking confirmation.

## Homework Statement

I'd like to show, if possible, that rotational invariance about some axis implies that angular momentum about that axis is conserved without using the Lagrangian formalism or Noether's theorem. The only proofs I have been able to find use a Lagrangian approach and I'm wondering if it's possible to do this using only geometry and vector mechanics.

## Homework Equations

Definition of vector angular momentum and torque, basic vector geometry in coordinate-free form.

## The Attempt at a Solution

Suppose that the interactions of a system S with the outside world can be described in terms of a potential energy function U, and that we measure U at some point P with position vector r with respect to some coordinate system embedded in S. Suppose that S is rotated by an arbitrarily small angle Δθ about one of the axes (Δθ points in the direction of the axis). Then the position vector of point P in this new coordinate system is r+Δθr. Suppose that this has not changed the measured value of U so that U(r) - U( r+Δθr) =0. This is true for an arbitrarily small rotation and therefore $$\lim_{\mid \Delta \vec{\theta} \times \vec{r} \mid \to 0} \frac{U(\vec{r} +\Delta {\vec{\theta}} \times \vec{r}) - U(\vec{r})}{\mid \Delta \vec{\theta} \times \vec{r} \mid} = 0$$

But this is simply the directional derivative of U in the direction of Δθr, which means that the force at point P in that direction is 0. By Newton's third law, this means that the torque about the axis of rotation experienced by S is also 0. Therefore angular momentum about that axis is constant in any motion of this system.

There's no specific thing here that I'm not sure of, I just have this nagging feeling that it shouldn't be this simple.

Bump?