# Conceptual problem with angular momentum conservation

1. Feb 4, 2015

### facenian

Let $L(\vec{r},\dot{\vec{r}})$ be the lagrangian function, usually to get angular momentum conservation one impose $\delta L=0$ and form there we get $\sum \vec{r}\wedge m\vec{v}=const$. There is however a conceptual problem with this procedure related to the fact that invariance under rotation does not necesarily means $\delta L=0$ but more generally $\delta L=\frac{d F(\vec{r},t)}{dt}$ and this gives
$$\sum \vec{r}\wedge m\vec{v}=F(\vec{r},t)$$
It seems that for a general lagrangian we can't derive angular momentum conservation unless additional hypotheses are introduced concerning the form of the lagrangian function. For instance if we work with the concrete lagrangian $L=\sum \frac{1}{2}mv^2-U(\vec{r},\dots)$ we do get $F(\vec{r},t)=const$, however for a general lagrangian, as I said before, we don't know. All this also applies to special relativity mechanics.
I would appreciate any comments.

2. Feb 4, 2015

### Staff: Mentor

Angular momentum is conserved in closed systems only if all external forces have a radial symmetry.

3. Feb 5, 2015

### facenian

Yes, and this amounts to imposing additional conditions on the lagrangian function, so my question remains unanswered. Is it possible to obtain angular momentum conservation from the symmetry principle of Lorentz(or Galilean) invariance without any additional condition on the lagrangian?

The problem with the usual derivation stems from imposing $\delta L=0$ instead of the more general invariance condition $\delta L=\frac{d F}{dt}$

4. Feb 5, 2015

### Staff: Mentor

The Lagrangian has to be invariant under rotation (otherwise your system does not follow the right symmetry and there is no reason to assume conserved angular momentum), which is exactly the condition given above.

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