# Prove rotational invariance leads to conservation of ang. momentum

1. Apr 16, 2013

### xWaffle

1. The problem statement, all variables and given/known data

Noether's theorem asserts a connection between invariance principles and conservation laws. In section 7.8 we saw that translational invariance of the Lagrangian implies conservation of total linear momentum. Here you will prove that rotational invariance of L implies conservation of total angular momentum. Suppose that the Lagrangian of an N-particle system is unchanged by rotations about a certain symmetry axis. (a) Without loss of generality, take this axis to be the z axis, and show that the Lagrangian is unchanged when all of the particles are simultaneously moved from (rα, θα, ϕα) to(rα, θα, ϕα + ϵ) (same ϵ for all particles). Hence show that

$\sum_{\alpha=1}^{N} \frac{\partial \mathcal{L}}{\partial \phi_{\alpha}} = 0$

(b) Use Lagrange's equations to show that this implies that the total angular momentum Lz about the symmetry axis is constant. In particular, if the Lagrangian is invariant under rotations about all axes, then all components of L are conserved.

2. Relevant equations

Lagrange Equations: Translational Invariance Example

No external force.

Simply, the change in potential is:
$\delta U = U(\overline{r}_{1} + d\overline{r}_{1}, \overline{r}_{2} + d\overline{r}_{2}, ... , t) - U(\overline{r}_{1}, \overline{r}_{2}, ... , t)$
$U = \sum_{\alpha\beta} U_{\alpha\beta}(\overline{r}_{\alpha} - \overline{r}_{\beta})$

And very similarly for the Lagrangian:
$\delta\mathcal{L} = \mathcal{L}(\overline{r}_{1} + d\overline{r}_{1}, \overline{r}_{2} + d\overline{r}_{2}, ... , t) - \mathcal{L}(\overline{r}_{1}, \overline{r}_{2}, ... , t)$

Stationary requires this change in action to be zero:
$\delta\mathcal{L} = 0$

And, earlier in the semester, we proved that a conservative force can be written as the negative gradient of a scalar valued function. We will use this here:
$\delta\mathcal{L} = d\overline{r} \bullet \nabla_{1} \mathcal{L} + d\overline{r} \bullet \nabla_{2} \mathcal{L} + ... = 0$

And to redefine some of the terms in the last line:
$d \overline{r} = \hat{x} dx$
$d \overline{r} \bullet \nabla_{1} \mathcal{L} = dx \frac{ \partial \mathcal{L} }{ \partial x_{1}}$

Now, plug them in to make it look a little bit more like what we want..
$dx \sum_{\alpha} \frac{ \partial \mathcal{L} }{ \partial x_{\alpha}} = 0$
$\sum_{\alpha} \frac{\partial \mathcal{L}}{\partial x_{\alpha}} = 0$

Lagrange Equations for any particle in the system:
$\frac{\partial \mathcal{L}}{\partial x_{\alpha}} = \frac{d}{dt} \left(\frac{\partial \mathcal{L}}{\partial \dot{x}_{\alpha}} \right)$

Which is the same as saying:
$\frac{\partial \mathcal{L}}{\partial x_{\alpha}} = \frac{d}{dt} \left( p_{x\alpha} \right)$
$\frac{\partial \mathcal{L}}{\partial x_{\alpha}} = \dot{p}_{x\alpha}$

And from here, out conclusion can be shown very simply:
$\sum_{\alpha} \dot{p}_{x\alpha} = \dot{p}_{total} = 0$
$\dot{p}_{total} = 0$
$p_{total} = 0 \rightarrow consant$

3. The attempt at a solution

We've been shown an example with translational invariance leading to conservation of momentum. He used the Lagrange equations and related them to U(r) potential energy, showed that with no external forces, a change in the Lagrangian can be written as a re-written form of the difference of potentials. Which then led to the full time derivative of the Lagrangian not explicitly depending on x (only x-dot), which meant p (momentum) was constant.

I've tried to approach it in a similar way, but I'm stuck trying to wrap my head around what would be the 'rotational equivalent' of his first step. Is this the wrong way to approach it?

Last edited: Apr 16, 2013